GSEB Class 11 Maths Solutions Chapter 9 Sequences and Series Exercise 9.2

Question 1. Find the sum of odd integers from 1 to 2001.
Answer: We need to find the sum: \( 1 + 3 + 5 + \dots + 2001 \).
This is an arithmetic progression (A.P.) where the first term is \( a = 1 \) and the common difference is \( d = 2 \).
Let \( T_n \) be the final term, so \( T_n = 2001 \).
Using the formula for the \( n^{th} \) term: \( T_n = a + (n - 1)d \).
\( \implies 2001 = 1 + (n - 1)2 \)
\( \implies 2000 = (n - 1)2 \)
\( \implies \frac{2000}{2} = n - 1 \)
\( \implies 1000 = n - 1 \)
\( \implies n = 1001 \).
Now, we calculate the sum of \( n \) terms using the formula \( S_n = \frac{n}{2} [2a + (n - 1)d] \).
\( S_{1001} = \frac{1001}{2} [2 \times 1 + (1001 - 1) \times 2] \)
\( \implies S_{1001} = \frac{1001}{2} [2 + 1000 \times 2] \)
\( \implies S_{1001} = \frac{1001}{2} [2 + 2000] \)
\( \implies S_{1001} = \frac{1001}{2} [2002] \)
\( \implies S_{1001} = 1001 \times 1001 \)
\( \implies S_{1001} = 1002001 \).
In simple words: We found how many odd numbers are there between 1 and 2001. Then, we used the sum formula for arithmetic progressions to add them all up.

Exam Tip: For problems involving sums of arithmetic progressions, always clearly identify the first term (a), common difference (d), and the number of terms (n) first. Then apply the relevant sum formula, \( S_n = \frac{n}{2}(a + T_n) \) or \( S_n = \frac{n}{2}[2a + (n-1)d] \).

Question 2. Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5?
Answer: We need to find the sum of natural numbers between 100 and 1000 that are multiples of 5.
The sequence starts with the first multiple of 5 greater than 100, which is \( 105 \).
The sequence ends with the last multiple of 5 less than 1000, which is \( 995 \).
So, the series is: \( 105, 110, 115, \dots, 995 \).
This is an arithmetic progression (A.P.) with first term \( a = 105 \) and common difference \( d = 5 \).
Let \( T_n \) be the last term, so \( T_n = 995 \).
Using the formula for the \( n^{th} \) term: \( T_n = a + (n - 1)d \).
\( \implies 995 = 105 + (n - 1)5 \)
\( \implies 995 - 105 = (n - 1)5 \)
\( \implies 890 = (n - 1)5 \)
\( \implies \frac{890}{5} = n - 1 \)
\( \implies 178 = n - 1 \)
\( \implies n = 179 \).
Now, we find the sum of these \( 179 \) terms using the formula \( S_n = \frac{n}{2} (a + T_n) \).
\( S_{179} = \frac{179}{2} (105 + 995) \)
\( \implies S_{179} = \frac{179}{2} (1100) \)<
\( \implies S_{179} = 179 \times 550 \)
\( \implies S_{179} = 98450 \).
In simple words: We first found the first and last numbers divisible by 5 between 100 and 1000. Then, we counted how many such numbers exist. Finally, we added all those numbers together using the sum formula for an A.P.

Exam Tip: When finding sums of numbers between two limits, be careful to include or exclude the limits based on whether the question says "between" or "from...to...". Always ensure the first and last terms are correct for your chosen common difference.

Question 3. In an A.P., the first term is 2 and the sum of first five terms is one-fourth of the sum of next five terms. Show that its 20th term is -112.
Answer: Let \( a \) be the first term and \( d \) be the common difference. We are given \( a = 2 \).
The sum of the first \( n \) terms of an A.P. is \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
Sum of the first five terms \( S_5 = \frac{5}{2}[2a + (5 - 1)d] \)
\( \implies S_5 = \frac{5}{2}[2(2) + 4d] \)
\( \implies S_5 = \frac{5}{2}[4 + 4d] \)
\( \implies S_5 = 5(2 + 2d) \).
The sum of the next five terms (i.e., from \( T_6 \) to \( T_{10} \)) can be found by taking the sum of 5 terms starting with \( T_6 \).
The sixth term \( T_6 = a + 5d \). We can treat this as the new first term for this five-term sequence.
Sum of next five terms = \( S'_5 = \frac{5}{2}[2(a+5d) + (5 - 1)d] \)
\( \implies S'_5 = \frac{5}{2}[2a + 10d + 4d] \)
\( \implies S'_5 = \frac{5}{2}[2a + 14d] \)
\( \implies S'_5 = 5(a + 7d) \).
Since \( a = 2 \),
\( \implies S'_5 = 5(2 + 7d) \).
According to the problem statement, the sum of first five terms is one-fourth of the sum of next five terms:
\( S_5 = \frac{1}{4} S'_5 \)
\( \implies 5(2 + 2d) = \frac{1}{4} \times 5(2 + 7d) \)
Dividing both sides by 5 and multiplying by 4:
\( \implies 4(2 + 2d) = 2 + 7d \)
\( \implies 8 + 8d = 2 + 7d \)
\( \implies 8d - 7d = 2 - 8 \)
\( \implies d = -6 \).
Now, we need to find the \( 20^{th} \) term, \( T_{20} \).
\( T_{20} = a + (20 - 1)d \)
\( \implies T_{20} = 2 + (19)(-6) \)
\( \implies T_{20} = 2 - 114 \)
\( \implies T_{20} = -112 \).
Thus, the \( 20^{th} \) term is -112, as required to show.
In simple words: We used the given information about the sum of the first five terms and the next five terms to find the common difference. Once we had the common difference, we used it to calculate the 20th term of the sequence.

Exam Tip: For problems involving sums of consecutive groups of terms in an A.P., remember that you can either use \( S_n \) and \( S_m \) to find \( S_m - S_n \) for terms between \( n+1 \) and \( m \), or treat the first term of the sub-sequence as a new 'a' value and apply the sum formula.

Question 4. How many terms of the A.P. -6, \( -\frac{11}{2} \), -5, ... are needed to give the sum -25?
Answer: Given the arithmetic progression (A.P.): \( -6, -\frac{11}{2}, -5, \dots \).
The first term is \( a = -6 \).
The common difference \( d = -\frac{11}{2} - (-6) = -\frac{11}{2} + 6 = \frac{-11 + 12}{2} = \frac{1}{2} \).
Let \( n \) be the number of terms. We are given that the sum of \( n \) terms \( S_n = -25 \).
Using the formula for the sum of \( n \) terms: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( \implies -25 = \frac{n}{2} [2(-6) + (n - 1)\frac{1}{2}] \)
\( \implies -50 = n [-12 + \frac{n - 1}{2}] \)
\( \implies -50 = n [\frac{-24 + n - 1}{2}] \)
\( \implies -100 = n(n - 25) \)
\( \implies -100 = n^2 - 25n \)
\( \implies n^2 - 25n + 100 = 0 \).
We can factor this quadratic equation:
\( \implies (n - 5)(n - 20) = 0 \).
This gives two possible values for \( n \):
\( \implies n = 5 \) or \( n = 20 \).
Both values of \( n \) result in the required sum of -25. This is possible if some terms are positive and some are negative, leading to the same sum for a different number of terms.
In simple words: We identified the starting term and the gap between terms in the sequence. Then, we used the sum formula for an A.P. to set up an equation. Solving this equation gave us two possible counts for the number of terms that would add up to -25.

Exam Tip: When solving for the number of terms \( n \) in an A.P. sum problem, you might get a quadratic equation. Always check both solutions. If \( n \) must be a positive integer, discard any negative or fractional answers. Sometimes, both positive integer solutions are valid.

Question 5. In an A.P., if the \( p^{th} \) term is \( \frac{1}{q} \) and the \( q^{th} \) term is \( \frac{1}{p} \), prove that the sum of first \( pq \) terms is \( \frac{1}{2}(pq + 1) \), where \( p \neq q \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We are given:
The \( p^{th} \) term, \( T_p = \frac{1}{q} \). So, \( a + (p - 1)d = \frac{1}{q} \dots (1) \)
The \( q^{th} \) term, \( T_q = \frac{1}{p} \). So, \( a + (q - 1)d = \frac{1}{p} \dots (2) \)
Subtracting equation (2) from equation (1):
\( [a + (p - 1)d] - [a + (q - 1)d] = \frac{1}{q} - \frac{1}{p} \)
\( \implies (p - 1 - q + 1)d = \frac{p - q}{pq} \)
\( \implies (p - q)d = \frac{p - q}{pq} \).
Since \( p \neq q \), we can divide both sides by \( (p - q) \):
\( \implies d = \frac{1}{pq} \).
Substitute the value of \( d \) into equation (1) to find \( a \):
\( a + (p - 1)\frac{1}{pq} = \frac{1}{q} \)
\( \implies a = \frac{1}{q} - \frac{p - 1}{pq} \)
\( \implies a = \frac{p - (p - 1)}{pq} \)
\( \implies a = \frac{p - p + 1}{pq} \)
\( \implies a = \frac{1}{pq} \).
Now, we need to find the sum of the first \( pq \) terms, \( S_{pq} \).
Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( S_{pq} = \frac{pq}{2} [2a + (pq - 1)d] \)
Substitute the values of \( a = \frac{1}{pq} \) and \( d = \frac{1}{pq} \):
\( \implies S_{pq} = \frac{pq}{2} [2 \left(\frac{1}{pq}\right) + (pq - 1)\left(\frac{1}{pq}\right)] \)
\( \implies S_{pq} = \frac{pq}{2} [\frac{2}{pq} + \frac{pq - 1}{pq}] \)
\( \implies S_{pq} = \frac{pq}{2} [\frac{2 + pq - 1}{pq}] \)
\( \implies S_{pq} = \frac{pq}{2} [\frac{pq + 1}{pq}] \)
\( \implies S_{pq} = \frac{1}{2} (pq + 1) \).
Hence, it is proved that the sum of the first \( pq \) terms is \( \frac{1}{2}(pq + 1) \).
In simple words: We used the given p-th and q-th terms to find the first term and common difference of the A.P. Once these values were known, we substituted them into the formula for the sum of an arithmetic progression to show the required result.

Exam Tip: For problems involving \( p^{th} \) and \( q^{th} \) terms, solving simultaneous equations for \( a \) and \( d \) is usually the first step. Algebraic manipulation and simplification are key to reaching the final proof.

Question 6. If the sum of a certain number of terms of the A.P.; 25, 22, 19, ... is 116. Find the last term?
Answer: Given the arithmetic progression (A.P.): \( 25, 22, 19, \dots \).
The first term is \( a = 25 \).
The common difference \( d = 22 - 25 = -3 \).
Let \( n \) be the number of terms. We are given that the sum of \( n \) terms \( S_n = 116 \).
Using the formula for the sum of \( n \) terms: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( \implies 116 = \frac{n}{2}[2(25) + (n - 1)(-3)] \)
\( \implies 232 = n[50 - 3n + 3] \)
\( \implies 232 = n[53 - 3n] \)
\( \implies 232 = 53n - 3n^2 \)
\( \implies 3n^2 - 53n + 232 = 0 \).
We can factor this quadratic equation by finding two numbers that multiply to \( 3 \times 232 = 696 \) and add up to \( -53 \). These numbers are -24 and -29.
\( \implies 3n^2 - 24n - 29n + 232 = 0 \)
\( \implies 3n(n - 8) - 29(n - 8) = 0 \)
\( \implies (n - 8)(3n - 29) = 0 \).
This gives two possible values for \( n \):
\( \implies n = 8 \) or \( n = \frac{29}{3} \).
Since the number of terms \( n \) must be a natural number, \( n \neq \frac{29}{3} \).
Therefore, \( n = 8 \).
The question asks for the last term, which is the \( 8^{th} \) term, \( T_8 \).
Using the formula \( T_n = a + (n - 1)d \):
\( T_8 = 25 + (8 - 1)(-3) \)
\( \implies T_8 = 25 + 7(-3) \)
\( \implies T_8 = 25 - 21 \)
\( \implies T_8 = 4 \).
Hence, the last term is \( 4 \).
In simple words: We used the given first term, common difference, and total sum to find the number of terms. Once we found the number of terms, we used it to calculate the value of the last term in the sequence.

Exam Tip: Always verify that the calculated number of terms (\( n \)) is a positive integer. If a quadratic equation yields two positive integer solutions for \( n \), both may be valid. For this problem, \( n \) must be a whole number for a 'last term' to exist in the usual sense.

Question 7. Find the sum of n terms of an A.P., whose \( k^{th} \) term is \( 5k + 1 \).
Answer: The \( k^{th} \) term of the A.P. is given by \( T_k = 5k + 1 \).
To find the first term (\( a \)), substitute \( k = 1 \):
\( T_1 = a = 5(1) + 1 = 6 \).
To find the second term (\( T_2 \)), substitute \( k = 2 \):
\( T_2 = 5(2) + 1 = 11 \).
The common difference (\( d \)) is the difference between consecutive terms:
\( d = T_2 - T_1 = 11 - 6 = 5 \).
So, we have the first term \( a = 6 \) and the common difference \( d = 5 \).
Now, we need to find the sum of \( n \) terms, \( S_n \).
Using the formula \( S_n = \frac{n}{2}[2a + (n - 1)d] \):
\( S_n = \frac{n}{2}[2(6) + (n - 1)5] \)
\( \implies S_n = \frac{n}{2}[12 + 5n - 5] \)
\( \implies S_n = \frac{n}{2}[5n + 7] \).
Thus, the sum of \( n \) terms of the A.P. is \( \frac{n(5n+7)}{2} \).
In simple words: We used the formula for the k-th term to find the first and second terms of the arithmetic progression. From these, we calculated the common difference. Finally, we used these values in the sum formula for an A.P. to express the sum of 'n' terms.

Exam Tip: When the \( k^{th} \) term is given as a linear expression in \( k \), like \( T_k = Ak + B \), the common difference is simply the coefficient of \( k \) (A). This can save time in verifying your calculations for \( d \).

Question 8. If the sum to \( n \) terms of an A.P. is \( (pn + qn^2) \), where \( p \) and \( q \) are constants, find the common difference?
Answer: Let \( S_n \) denote the sum of \( n \) terms of the A.P.
We are given that \( S_n = pn + qn^2 \).
To find the first term (\( T_1 \)), we set \( n = 1 \):
\( S_1 = T_1 = p(1) + q(1)^2 = p + q \).
To find the sum of the first two terms (\( S_2 \)), we set \( n = 2 \):
\( S_2 = p(2) + q(2)^2 = 2p + 4q \).
The second term (\( T_2 \)) can be found by subtracting the sum of the first term from the sum of the first two terms:
\( T_2 = S_2 - S_1 \)
\( \implies T_2 = (2p + 4q) - (p + q) \)
\( \implies T_2 = 2p - p + 4q - q \)
\( \implies T_2 = p + 3q \).
The common difference (\( d \)) is the difference between the second term and the first term:
\( d = T_2 - T_1 \)
\( \implies d = (p + 3q) - (p + q) \)
\( \implies d = p - p + 3q - q \)
\( \implies d = 2q \).
Thus, the common difference of this A.P. is \( 2q \).
In simple words: We used the given formula for the sum of 'n' terms to find the first and second terms of the sequence. By subtracting the first term from the second, we directly found the common difference of the arithmetic progression.

Exam Tip: Remember that \( T_n = S_n - S_{n-1} \). This relationship is very useful for finding the \( n^{th} \) term when the sum formula is given. For \( T_1 \), \( S_1 \) is simply \( T_1 \).

Question 9. The sums of first \( n \) terms of two arithmetic progressions are in the ratio \( (5n + 4) : (9n + 6) \). Find the ratio of their \( 18^{th} \) terms?
Answer: Let \( a_1 \) and \( d_1 \) be the first term and common difference of the first A.P., respectively. Let \( a_2 \) and \( d_2 \) be the first term and common difference of the second A.P., respectively.
Let \( S_{1n} \) and \( S_{2n} \) be their respective sums of \( n \) terms.
The ratio of their sums is given as:
\( \frac{S_{1n}}{S_{2n}} = \frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} \).
We are given that \( \frac{S_{1n}}{S_{2n}} = \frac{5n + 4}{9n + 6} \).
So, \( \frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{5n + 4}{9n + 6} \dots (1) \).
We need to find the ratio of their \( 18^{th} \) terms, \( \frac{T_{1,18}}{T_{2,18}} \).
The \( 18^{th} \) term of an A.P. is given by \( T_{18} = a + (18 - 1)d = a + 17d \).
So, the required ratio is \( \frac{a_1 + 17d_1}{a_2 + 17d_2} \).
To match the form of equation (1), we multiply the numerator and denominator by 2:
\( \frac{a_1 + 17d_1}{a_2 + 17d_2} = \frac{2(a_1 + 17d_1)}{2(a_2 + 17d_2)} = \frac{2a_1 + 34d_1}{2a_2 + 34d_2} \dots (2) \).
Comparing equation (1) and equation (2), we see that the term \( (n - 1)d \) in the sum formula corresponds to \( 34d \) in the term formula. This implies that \( n - 1 = 34 \).
\( \implies n = 35 \).
Now, substitute \( n = 35 \) into the given ratio of sums (equation 1):
\( \frac{2a_1 + (35 - 1)d_1}{2a_2 + (35 - 1)d_2} = \frac{5(35) + 4}{9(35) + 6} \)
\( \implies \frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{175 + 4}{315 + 6} \)
\( \implies \frac{2a_1 + 34d_1}{2a_2 + 34d_2} = \frac{179}{321} \).
From equation (2), we know that \( \frac{2a_1 + 34d_1}{2a_2 + 34d_2} \) is the ratio of the \( 18^{th} \) terms.
Therefore, the ratio of their \( 18^{th} \) terms is \( \frac{179}{321} \).
In simple words: We used the given ratio of sums of 'n' terms for two A.P.s. To find the ratio of their 18th terms, we found what value of 'n' would make the sum formula look like the term formula. Using that 'n' value in the given ratio of sums, we got the answer.

Exam Tip: A key trick for this type of problem is to equate the `(n-1)` in the sum formula `(2a + (n-1)d)` to `2(k-1)` for the `k^{th}` term, where `k` is the term number you want to find. This means `n = 2k-1`.

Question 10. If the sum of first \( p \) terms of an A.P. is equal to the sum of first \( q \) terms, then find the sum of first \( (p + q) \) terms?
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The sum of \( n \) terms of an A.P. is given by \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
Given that the sum of the first \( p \) terms is equal to the sum of the first \( q \) terms:
\( S_p = S_q \)
\( \implies \frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d] \).
Multiplying both sides by 2 and expanding:
\( \implies p[2a + (p - 1)d] = q[2a + (q - 1)d] \)
\( \implies 2ap + p(p - 1)d = 2aq + q(q - 1)d \)
Rearrange terms to one side:
\( \implies 2ap - 2aq + p(p - 1)d - q(q - 1)d = 0 \)
Factor out \( 2a \) and \( d \):
\( \implies 2a(p - q) + d[p(p - 1) - q(q - 1)] = 0 \)
Expand the terms inside the square brackets:
\( \implies 2a(p - q) + d[p^2 - p - (q^2 - q)] = 0 \)
\( \implies 2a(p - q) + d[p^2 - q^2 - p + q] = 0 \)
Factor \( (p^2 - q^2) \) as \( (p - q)(p + q) \) and \( (-p + q) \) as \( -(p - q) \):
\( \implies 2a(p - q) + d[(p - q)(p + q) - (p - q)] = 0 \)
Factor out \( (p - q) \) from the \( d \) term:
\( \implies 2a(p - q) + d(p - q)[(p + q) - 1] = 0 \)
Now, factor out \( (p - q) \) from the entire expression:
\( \implies (p - q)[2a + d(p + q - 1)] = 0 \).
Since \( p \neq q \) (if \( p = q \), the statement is trivially true and not a useful problem), we must have \( (p - q) \neq 0 \).
Therefore, the other factor must be zero:
\( 2a + d(p + q - 1) = 0 \dots (1) \).
Now, we need to find the sum of the first \( (p + q) \) terms, \( S_{p+q} \).
Using the sum formula:
\( S_{p+q} = \frac{p+q}{2}[2a + ((p + q) - 1)d] \).
From equation (1), we know that \( 2a + (p + q - 1)d = 0 \).
Substitute this into the expression for \( S_{p+q} \):
\( \implies S_{p+q} = \frac{p+q}{2}[0] \)
\( \implies S_{p+q} = 0 \).
Thus, the sum of the first \( (p + q) \) terms is \( 0 \).
In simple words: We set the sum of 'p' terms equal to the sum of 'q' terms and simplified the equation. This led us to discover that a certain part of the sum formula for 'p+q' terms must be zero. Using this, we found the sum of the first 'p+q' terms.

Exam Tip: This is a common and important result. When \( S_p = S_q \) and \( p \neq q \), it always implies that \( S_{p+q} = 0 \). Recognizing this pattern can help you quickly confirm your derived equation `2a + (p + q - 1)d = 0`.

Question 11. Sum of the first \( p, q \) and \( r \) terms of an A.P. are \( a, b \) and \( c \) respectively. Prove that \( \frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q) = 0 \).
Answer: Let \( A \) be the first term and \( D \) be the common difference of the A.P.
The sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2}[2A + (n - 1)D] \).
Given:
Sum of \( p \) terms = \( a \): \( S_p = a \)
\( \implies \frac{p}{2}[2A + (p - 1)D] = a \)
Dividing by \( p \):
\( \implies A + \frac{1}{2}(p - 1)D = \frac{a}{p} \dots (1) \).
Sum of \( q \) terms = \( b \): \( S_q = b \)
\( \implies \frac{q}{2}[2A + (q - 1)D] = b \)
Dividing by \( q \):
\( \implies A + \frac{1}{2}(q - 1)D = \frac{b}{q} \dots (2) \).
Sum of \( r \) terms = \( c \): \( S_r = c \)
\( \implies \frac{r}{2}[2A + (r - 1)D] = c \)
Dividing by \( r \):
\( \implies A + \frac{1}{2}(r - 1)D = \frac{c}{r} \dots (3) \).
Now, we multiply equation (1) by \( (q - r) \), equation (2) by \( (r - p) \), and equation (3) by \( (p - q) \). Then, we add the three resulting equations.
\( \frac{a}{p}(q - r) = A(q - r) + \frac{1}{2}(p - 1)D(q - r) \)
\( \frac{b}{q}(r - p) = A(r - p) + \frac{1}{2}(q - 1)D(r - p) \)
\( \frac{c}{r}(p - q) = A(p - q) + \frac{1}{2}(r - 1)D(p - q) \).
Adding these three expressions:
LHS: \( \frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q) \).
RHS: \( A(q - r + r - p + p - q) + \frac{1}{2}D[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)] \).
Consider the coefficient of \( A \):
\( q - r + r - p + p - q = 0 \).
Consider the coefficient of \( \frac{1}{2}D \):
Let \( K = (p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q) \).
\( K = (pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q) \).
When we expand and collect terms:
\( pq \) and \( -qp \) cancel.
\( -pr \) and \( rp \) cancel.
\( -q \) and \( q \) cancel.
\( r \) and \( -r \) cancel.
\( qr \) and \( -rq \) cancel.
\( p \) and \( -p \) cancel.
Thus, \( K = 0 \).
So, the RHS becomes \( A(0) + \frac{1}{2}D(0) = 0 \).
Therefore, \( \frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q) = 0 \).
This completes the proof.
In simple words: We wrote out the sum formulas for p, q, and r terms, then divided each by its term count. By multiplying each of these new equations by specific differences (like q-r) and adding them up, all the terms involving the first term and common difference canceled out, leaving the required result of zero.

Exam Tip: This type of proof often relies on a cyclic summation (e.g., \( (x-y) + (y-z) + (z-x) = 0 \)). Pay close attention to how terms cancel out when expressions are expanded and combined. Be very careful with signs.

Question 12. The ratio of the sums of first \( m \) and \( n \) terms of an A.P. is \( m^2 : n^2 \). Show that the ratio of \( m^{th} \) and \( n^{th} \) terms is \( (2m - 1) : (2n - 1) \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
The sum of \( k \) terms of an A.P. is \( S_k = \frac{k}{2}[2a + (k - 1)d] \).
Given the ratio of the sums of the first \( m \) and \( n \) terms:
\( \frac{S_m}{S_n} = \frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2} \).
Simplifying, we get:
\( \frac{m[2a + (m - 1)d]}{n[2a + (n - 1)d]} = \frac{m^2}{n^2} \).
Dividing both sides by \( \frac{m}{n} \):
\( \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \dots (1) \).
We want to find the ratio of the \( m^{th} \) and \( n^{th} \) terms: \( \frac{T_m}{T_n} = \frac{a + (m - 1)d}{a + (n - 1)d} \).
To convert the expression \( \frac{2a + (k - 1)d}{2a + (j - 1)d} \) into the form \( \frac{a + (k' - 1)d}{a + (j' - 1)d} \), we can use a standard technique.
Specifically, to obtain \( a + (m - 1)d \) from \( 2a + (X)d \), we need \( X = 2(m - 1) \).
So, in equation (1), we replace \( m \) with \( (2m - 1) \) and \( n \) with \( (2n - 1) \).
Substituting \( m \rightarrow (2m - 1) \) and \( n \rightarrow (2n - 1) \) into equation (1):
\( \frac{2a + ((2m - 1) - 1)d}{2a + ((2n - 1) - 1)d} = \frac{2m - 1}{2n - 1} \).
\( \implies \frac{2a + (2m - 2)d}{2a + (2n - 2)d} = \frac{2m - 1}{2n - 1} \).
Factor out 2 from the numerator and denominator on the LHS:
\( \implies \frac{2[a + (m - 1)d]}{2[a + (n - 1)d]} = \frac{2m - 1}{2n - 1} \).
\( \implies \frac{a + (m - 1)d}{a + (n - 1)d} = \frac{2m - 1}{2n - 1} \).
Since \( T_m = a + (m - 1)d \) and \( T_n = a + (n - 1)d \), this gives:
\( \frac{T_m}{T_n} = \frac{2m - 1}{2n - 1} \).
Thus, the ratio of the \( m^{th} \) and \( n^{th} \) terms is \( (2m - 1) : (2n - 1) \), as required to show.
In simple words: We began with the given ratio of sums of 'm' and 'n' terms. By replacing 'm' and 'n' in this sum ratio with specific values, we could transform the sum expressions into term expressions. This allowed us to directly find the ratio of the m-th and n-th terms.

Exam Tip: This is a classic result in arithmetic progressions. Remember that if \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \), then \( \frac{T_m}{T_n} = \frac{2m - 1}{2n - 1} \). This relationship allows for quick verification of answers.

Question 13. If the sum of first \( n \) terms of an A.P. is \( 3n^2 + 5n \) and its \( m^{th} \) term is 164, find the value of \( m \)?
Answer: Let \( S_n \) represent the sum of the first \( n \) terms of the A.P.
We are given that \( S_n = 3n^2 + 5n \).
To find the first term (\( a \) or \( T_1 \)), we set \( n = 1 \):
\( S_1 = T_1 = 3(1)^2 + 5(1) = 3 + 5 = 8 \).
To find the sum of the first two terms (\( S_2 \)), we set \( n = 2 \):
\( S_2 = 3(2)^2 + 5(2) = 3(4) + 10 = 12 + 10 = 22 \).
The second term (\( T_2 \)) can be found by subtracting the sum of the first term from the sum of the first two terms:
\( T_2 = S_2 - S_1 = 22 - 8 = 14 \).
The common difference (\( d \)) is the difference between the second term and the first term:
\( d = T_2 - T_1 = 14 - 8 = 6 \).
So, the first term \( a = 8 \) and the common difference \( d = 6 \).
We are given that the \( m^{th} \) term of this A.P. is 164.
Using the formula for the \( m^{th} \) term: \( T_m = a + (m - 1)d \).
\( \implies 164 = 8 + (m - 1)6 \)
\( \implies 164 - 8 = (m - 1)6 \)
\( \implies 156 = (m - 1)6 \)
\( \implies \frac{156}{6} = m - 1 \)
\( \implies 26 = m - 1 \)
\( \implies m = 26 + 1 \)
\( \implies m = 27 \).
Therefore, the value of \( m \) is \( 27 \).
In simple words: First, we used the sum formula to find the starting term and the common difference of the A.P. Then, using the formula for the m-th term and the given value of 164, we solved for 'm'.

Exam Tip: For problems where \( S_n \) is a quadratic in \( n \) (e.g., \( An^2 + Bn \)), the common difference is \( 2A \). In this case, \( A=3 \), so \( d = 2 \times 3 = 6 \), which matches our calculation and serves as a quick check.

Question 14. Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Answer: Let the five numbers to be inserted between 8 and 26 be \( A_1, A_2, A_3, A_4, A_5 \).
The resulting sequence will be an arithmetic progression:
\( 8, A_1, A_2, A_3, A_4, A_5, 26 \).
In this A.P., the first term is \( a = 8 \).
The last term is \( 26 \). There are \( 5 \) inserted terms plus the two given numbers, making a total of \( 5 + 2 = 7 \) terms.
So, \( T_7 = 26 \).
Using the formula for the \( n^{th} \) term of an A.P., \( T_n = a + (n - 1)d \), where \( d \) is the common difference:
\( T_7 = a + (7 - 1)d \)
\( \implies 26 = 8 + 6d \)
\( \implies 26 - 8 = 6d \)
\( \implies 18 = 6d \)
\( \implies d = \frac{18}{6} \)
\( \implies d = 3 \).
Now that we have the common difference, we can find the five inserted numbers:
\( A_1 = T_2 = a + d = 8 + 3 = 11 \)
\( A_2 = T_3 = a + 2d = 8 + 2(3) = 8 + 6 = 14 \)
\( A_3 = T_4 = a + 3d = 8 + 3(3) = 8 + 9 = 17 \)
\( A_4 = T_5 = a + 4d = 8 + 4(3) = 8 + 12 = 20 \)
\( A_5 = T_6 = a + 5d = 8 + 5(3) = 8 + 15 = 23 \).
The required five numbers are \( 11, 14, 17, 20, \) and \( 23 \).
In simple words: We created an arithmetic sequence with 8 as the first term and 26 as the seventh term. Then, we calculated the common difference between terms. Using this difference, we filled in the five missing numbers in the sequence.

Exam Tip: When inserting \( m \) numbers between two given numbers \( A \) and \( B \) to form an A.P., the total number of terms in the sequence becomes \( m+2 \). The common difference is then \( d = \frac{B-A}{m+1} \). This shortcut can save time.

Question 15. If \( \frac{a^n + b^n}{a^{n-1} + b^{n-1}} \) is the A.M. between \( a \) and \( b \), then find the value of \( n \)?
Answer: The arithmetic mean (A.M.) between two numbers \( a \) and \( b \) is given by \( \frac{a+b}{2} \).
According to the problem, we have:
\( \frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a+b}{2} \).
Cross-multiply the terms:
\( \implies 2(a^n + b^n) = (a + b)(a^{n-1} + b^{n-1}) \).
Expand the right side:
\( \implies 2a^n + 2b^n = a \cdot a^{n-1} + a \cdot b^{n-1} + b \cdot a^{n-1} + b \cdot b^{n-1} \)
\( \implies 2a^n + 2b^n = a^n + ab^{n-1} + a^{n-1}b + b^n \).
Rearrange all terms to one side:
\( \implies 2a^n - a^n + 2b^n - b^n - ab^{n-1} - a^{n-1}b = 0 \)
\( \implies a^n + b^n - ab^{n-1} - a^{n-1}b = 0 \).
Group terms for factoring:
\( \implies (a^n - a^{n-1}b) - (ab^{n-1} - b^n) = 0 \)
Factor out \( a^{n-1} \) from the first group and \( b^{n-1} \) from the second group:
\( \implies a^{n-1}(a - b) - b^{n-1}(a - b) = 0 \).
Now, factor out the common term \( (a - b) \):
\( \implies (a - b)(a^{n-1} - b^{n-1}) = 0 \).
For this product to be zero, either \( (a - b) = 0 \) or \( (a^{n-1} - b^{n-1}) = 0 \).
If \( a - b = 0 \), then \( a = b \). In this case, the A.M. is \( a \), and the given expression also becomes \( \frac{a^n + a^n}{a^{n-1} + a^{n-1}} = \frac{2a^n}{2a^{n-1}} = a \), so any \( n \) would be valid. However, typically in such problems, \( a \neq b \) is implied for a unique solution for \( n \).
Assuming \( a \neq b \), we must have:
\( a^{n-1} - b^{n-1} = 0 \)
\( \implies a^{n-1} = b^{n-1} \).
If \( b \neq 0 \), we can divide by \( b^{n-1} \):
\( \implies \frac{a^{n-1}}{b^{n-1}} = 1 \)
\( \implies \left(\frac{a}{b}\right)^{n-1} = 1 \).
For \( \left(\frac{a}{b}\right)^{n-1} = 1 \) to be true (and for \( a \neq b \)), the exponent \( n - 1 \) must be \( 0 \).
\( \implies n - 1 = 0 \)
\( \implies n = 1 \).
In simple words: We set the given complex expression for the mean equal to the simple formula for an arithmetic mean. By cross-multiplying and simplifying, we factored the equation. Assuming the two numbers are different, we found that the exponent 'n' must be 1.

Exam Tip: When faced with exponential equations of the form \( x^k = y^k \), if \( x \neq y \), then \( k \) must be 0. If \( x=y \), then any \( k \) is valid. Always consider the conditions under which a solution is unique.

Question 16. Between 1 and 31, \( m \) numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of \( 7^{th} \) and \( (m - 1)^{th} \) numbers is 5:9. Find the value of \( m \)?
Answer: Let \( A_1, A_2, \dots, A_m \) be the \( m \) arithmetic means inserted between 1 and 31.
The resulting arithmetic progression is \( 1, A_1, A_2, \dots, A_m, 31 \).
The first term of this A.P. is \( a = 1 \).
The total number of terms in this A.P. is \( m + 2 \).
The last term, \( T_{m+2} = 31 \).
Using the formula \( T_n = a + (n - 1)d \):
\( T_{m+2} = a + ((m + 2) - 1)d \)
\( \implies 31 = 1 + (m + 1)d \)
\( \implies 30 = (m + 1)d \)
\( \implies d = \frac{30}{m + 1} \).
Now, let's find the \( 7^{th} \) and \( (m - 1)^{th} \) inserted numbers.
The \( 7^{th} \) inserted number is \( A_7 \), which is the \( 8^{th} \) term of the A.P. (\( T_8 \)).
\( A_7 = T_8 = a + 7d = 1 + 7\left(\frac{30}{m + 1}\right) \)
\( \implies A_7 = 1 + \frac{210}{m + 1} = \frac{m + 1 + 210}{m + 1} = \frac{m + 211}{m + 1} \).
The \( (m - 1)^{th} \) inserted number is \( A_{m-1} \), which is the \( m^{th} \) term of the A.P. (\( T_m \)).
\( A_{m-1} = T_m = a + (m - 1)d = 1 + (m - 1)\left(\frac{30}{m + 1}\right) \)
\( \implies A_{m-1} = 1 + \frac{30(m - 1)}{m + 1} = \frac{m + 1 + 30m - 30}{m + 1} = \frac{31m - 29}{m + 1} \).
We are given that the ratio of the \( 7^{th} \) and \( (m - 1)^{th} \) numbers is 5:9.
\( \frac{A_7}{A_{m-1}} = \frac{5}{9} \)
\( \implies \frac{\frac{m + 211}{m + 1}}{\frac{31m - 29}{m + 1}} = \frac{5}{9} \)
\( \implies \frac{m + 211}{31m - 29} = \frac{5}{9} \).
Cross-multiply:
\( \implies 9(m + 211) = 5(31m - 29) \)
\( \implies 9m + 1899 = 155m - 145 \)
Group terms:
\( \implies 1899 + 145 = 155m - 9m \)
\( \implies 2044 = 146m \)
\( \implies m = \frac{2044}{146} \)
\( \implies m = 14 \).
Therefore, the value of \( m \) is \( 14 \).
In simple words: We set up an arithmetic progression with 'm' numbers inserted between 1 and 31. We found the common difference in terms of 'm'. Then, we wrote out the expressions for the 7th and (m-1)th inserted numbers. Using their given ratio, we solved the resulting equation for 'm'.

Exam Tip: Carefully distinguish between the \( k^{th} \) inserted number (\( A_k \)) and the \( k^{th} \) term of the overall sequence (\( T_k \)). \( A_k \) corresponds to \( T_{k+1} \). This distinction is vital for setting up the correct equations.

Question 17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs. 5 every month, what amount he will pay in the 30th instalment?
Answer: The loan instalments form an arithmetic progression (A.P.).
The first instalment (first term) is \( a = \text{Rs. } 100 \).
He increases the instalment by Rs. 5 every month, so the common difference is \( d = \text{Rs. } 5 \).
We need to find the amount he will pay in the \( 30^{th} \) instalment, which is the \( 30^{th} \) term of the A.P., \( T_{30} \).
Using the formula for the \( n^{th} \) term of an A.P., \( T_n = a + (n - 1)d \):
For \( n = 30 \):
\( T_{30} = 100 + (30 - 1)5 \)
\( \implies T_{30} = 100 + (29)5 \)
\( \implies T_{30} = 100 + 145 \)
\( \implies T_{30} = 245 \).
Therefore, the man will pay Rs. 245 in the \( 30^{th} \) instalment.
In simple words: We identified the starting payment as the first term and the monthly increase as the common difference of an A.P. Then, we used the formula for the n-th term to calculate the payment amount for the 30th instalment.

Exam Tip: Clearly identify what the question is asking for: total amount paid (sum of terms) or a specific instalment amount (a particular term). This will determine which formula (sum or \( n^{th} \) term) you need to use.

Question 18. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of sides of the polygon?
Answer: The interior angles of the polygon form an arithmetic progression (A.P.).
The smallest angle (first term) is \( a = 120^\circ \).
The difference between consecutive interior angles (common difference) is \( d = 5^\circ \).
Let the polygon have \( n \) sides. The number of angles in the polygon is also \( n \).
The sum of \( n \) terms of an A.P. is given by \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
Substituting the values of \( a \) and \( d \):
\( S_n = \frac{n}{2}[2(120^\circ) + (n - 1)5^\circ] \)
\( \implies S_n = \frac{n}{2}[240^\circ + 5n^\circ - 5^\circ] \)
\( \implies S_n = \frac{n}{2}[235^\circ + 5n^\circ] \).
We also know that the sum of the interior angles of an \( n \)-sided polygon is given by the formula \( S_n = (n - 2) \times 180^\circ \).
Equating the two expressions for \( S_n \):
\( \frac{n}{2}[235 + 5n] = (n - 2)180 \).
Multiply both sides by 2:
\( n(235 + 5n) = 2(n - 2)180 \)
\( \implies 235n + 5n^2 = 360n - 720 \).
Rearrange into a standard quadratic form:
\( \implies 5n^2 + 235n - 360n + 720 = 0 \)
\( \implies 5n^2 - 125n + 720 = 0 \).
Divide the entire equation by 5:
\( \implies n^2 - 25n + 144 = 0 \).
Factor the quadratic equation:
\( \implies (n - 9)(n - 16) = 0 \).
This gives two possible values for \( n \):
\( \implies n = 9 \) or \( n = 16 \).
Now, we must check if both values are valid. The angles of a convex polygon must be less than \( 180^\circ \). The largest angle would be \( T_n = a + (n - 1)d \).
Case 1: If \( n = 16 \).
\( T_{16} = 120^\circ + (16 - 1)5^\circ = 120^\circ + 15 \times 5^\circ = 120^\circ + 75^\circ = 195^\circ \).
An angle of \( 195^\circ \) is greater than \( 180^\circ \), which is not possible for a convex polygon. So, \( n = 16 \) is not a valid solution.
Case 2: If \( n = 9 \).
\( T_9 = 120^\circ + (9 - 1)5^\circ = 120^\circ + 8 \times 5^\circ = 120^\circ + 40^\circ = 160^\circ \).
All angles in this case are less than \( 180^\circ \). So, \( n = 9 \) is a valid solution.
Therefore, the number of sides of the polygon is \( 9 \).
In simple words: We set up two expressions for the sum of the polygon's angles: one using the A.P. formula and another using the polygon's property. Equating them gave a quadratic equation for the number of sides. We checked both possible solutions to ensure the largest angle was less than 180 degrees, finding the correct number of sides.

Exam Tip: When finding the number of sides of a polygon where angles form an A.P., always remember to check if any of the angles exceed or equal \( 180^\circ \). Such solutions are not valid for a convex polygon, as interior angles must be less than \( 180^\circ \).