CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set J

Question. If the nth term of a progression be a linear expression in n, then prove that this progression is an AP.
Answer: Let the nth term of a given progression be given by
\( T_n = an + b \), where \( a \) and \( b \) are constants.
Then, \( T_{n–1} = a(n – 1) + b = [(an + b) – a] \)
\( \therefore (T_n – T_{n–1}) = (an + b) – [(an + b) – a] = a \), which is a constant.
Hence, the given progression is an AP.

Write the first three terms in each of the sequences defined by the following -

Question. \( a_n = 3n + 2 \)
Answer: We have, \( a_n = 3n + 2 \). Putting \( n = 1, 2 \) and \( 3 \), we get
\( a_1 = 3 \times 1 + 2 = 3 + 2 = 5 \),
\( a_2 = 3 \times 2 + 2 = 6 + 2 = 8 \),
\( a_3 = 3 \times 3 + 2 = 9 + 2 = 11 \)
Thus, the required first three terms of the sequence defined by \( a_n = 3n + 2 \) are \( 5, 8 \), and \( 11 \).

Question. \( a_n = n^2 + 1 \)
Answer: We have, \( a_n = n^2 + 1 \). Putting \( n = 1, 2, \) and \( 3 \) we get
\( a_1 = 1^2 + 1 = 1 + 1 = 2 \)
\( a_2 = 2^2 + 1 = 4 + 1 = 5 \)
\( a_3 = 3^2 + 1 = 9 + 1 = 10 \)
Thus, the first three terms of the sequence defined by \( a_n = n^2 + 1 \) are \( 2, 5 \) and \( 10 \).

Question. Write the first five terms of the sequence defined by \( a_n = (–1)^{n–1} \cdot 2^n \)
Answer: \( a_n = (–1)^{n–1} \times 2^n \). Putting \( n = 1, 2, 3, 4, \) and \( 5 \) we get
\( a_1 = (–1)^{1–1} \times 2^1 = (–1)^0 \times 2 = 2 \)
\( a_2 = (–1)^{2–1} \times 2^2 = (–1)^1 \times 4 = – 4 \)
\( a_3 = (–1)^{3–1} \times 2^3 = (–1)^2 \times 8 = 8 \)
\( a_4 = (–1)^{4–1} \times 2^4 = (–1)^3 \times 16 = –16 \)
\( a_5 = (–1)^{5–1} \times 2^5 = (–1)^4 \times 32 = 32 \)
Thus the first five term of the sequence are \( 2, –4, 8, –16, 32 \).

Question. The nth term of a sequence is \( 3n – 2 \). Is the sequence an A.P. ? If so, find its 10th term.
Answer: We have \( a_n = 3n – 2 \). Clearly \( a_n \) is a linear expression in \( n \). So, the given sequence is an A.P. with common difference \( 3 \).
Putting \( n = 10 \), we get \( a_{10} = 3 \times 10 – 2 = 28 \)

REMARK : It is evident from the above examples that a sequence is not an A.P. if its nth term is not a linear expression in \( n \).

Question. Find the 12th, 24th and nth term of the A.P. given by \( 9, 13, 17, 21, 25, ……… \)
Answer: We have, \( a = \) First term \( = 9 \) and, \( d = \) Common difference \( = 4 \) [\( \because 13 – 9 = 4, 17 – 13 = 4, 21 – 17 = 4 \) etc.]
We know that the nth term of an A.P. with first term \( a \) and common difference \( d \) is given by \( a_n = a + (n – 1)d \)
Therefore, \( a_{12} = a + (12 – 1) d = a + 11d = 9 + 11 \times 4 = 53 \)
\( a_{24} = a + (24 – 1) d = a + 23 d = 9 + 23 \times 4 = 101 \)
and, \( a_n = a + (n – 1) d = 9 + (n – 1) \times 4 = 4n + 5 \)
\( a_{12} = 53, a_{24} = 101 \) and \( a_n = 4n + 5 \)

Question. Which term of the sequence \( –1, 3, 7, 11, ….. \), is \( 95 \)?
Answer: Clearly, the given sequence is an A.P. We have, \( a = \) first term \( = –1 \) and, \( d = \) Common difference \( = 4 \).
Let \( 95 \) be the \( n^{th} \) term of the given A.P. then, \( a_n = 95 \)
\( \Rightarrow a + (n – 1) d = 95 \)
\( \Rightarrow – 1 + (n – 1) \times 4 = 95 \)
\( \Rightarrow – 1 + 4n – 4 = 95 \Rightarrow 4n – 5 = 95 \)
\( \Rightarrow 4n = 100 \Rightarrow n = 25 \)
Thus, \( 95 \) is 25th term of the given sequence.

Question. Which term of the sequence \( 4, 9, 14, 19, …… \) is \( 124 \)?
Answer: Clearly, the given sequence is an A.P. with first term \( a = 4 \) and common difference \( d = 5 \).
Let \( 124 \) be the \( n^{th} \) term of the given sequence. Then, \( a_n = 124 \)
\( a + (n – 1) d = 124 \)
\( \Rightarrow 4 + (n – 1) \times 5 = 124 \)
\( \Rightarrow n = 25 \)
Hence, 25th term of the given sequence is \( 124 \).

Question. The 10th term of an A.P. is \( 52 \) and 16th term is \( 82 \). Find the 32nd term and the general term.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Let the A.P. be \( a_1, a_2, a_3, ….. a_n, …… \)
It is given that \( a_{10} = 52 \) and \( a_{16} = 82 \)
\( \Rightarrow a + (10 – 1) d = 52 \) and \( a + (16 – 1) d = 82 \)
\( \Rightarrow a + 9d = 52 \dots (i) \) and \( a + 15d = 82 \dots (ii) \)
Subtracting equation (ii) from equation (i), we get \( –6d = – 30 \Rightarrow d = 5 \)
Putting \( d = 5 \) in equation (i), we get \( a + 45 = 52 \Rightarrow a = 7 \)
\( \therefore a_{32} = a + (32 – 1) d = 7 + 31 \times 5 = 162 \)
and, \( a_n = a + (n – 1) d = 7 + (n – 1) \times 5 = 5n + 2 \).
Hence \( a_{32} = 162 \) and \( a_n = 5n + 2 \).

Question. Determine the general term of an A.P. whose \( 7^{th} \) term is \( –1 \) and \( 16^{th} \) term \( 17 \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Let the A.P. be \( a_1, a_2, a_3, \dots a_n, \dots \)
It is given that \( a_7 = – 1 \) and \( a_{16} = 17 \)
\( a + (7 – 1) d = – 1 \) and, \( a + (16 – 1) d = 17 \)
\( \Rightarrow a + 6d = – 1 \dots (i) \) and, \( a + 15d = 17 \dots (ii) \)
Subtracting equation (i) from equation (ii), we get \( 9d = 18 \Rightarrow d = 2 \)
Putting \( d = 2 \) in equation (i), we get \( a + 12 = – 1 \Rightarrow a = – 13 \)
Now, General term \( = a_n = a + (n – 1) d = – 13 + (n – 1) \times 2 = 2n – 15 \)

Question. If five times the fifth term of an A.P. is equal to 8 times its eighth term, show that its 13th term is zero.
Answer: Let \( a_1, a_2, a_3, \dots, a_n, \dots \) be the A.P. with its first term \( = a \) and common difference \( = d \).
It is given that \( 5a_5 = 8a_8 \)
\( \Rightarrow 5(a + 4d) = 8 (a + 7d) \)
\( \Rightarrow 5a + 20d = 8a + 56d \Rightarrow 3a + 36d = 0 \)
\( \Rightarrow 3(a + 12d) = 0 \Rightarrow a + 12d = 0 \)
\( \Rightarrow a + (13 – 1) d = 0 \Rightarrow a_{13} = 0 \)

Question. If the mth term of an A.P. be \( 1/n \) and nth term be \( 1/m \), then show that its \( (mn)^{th} \) term is \( 1 \).
Answer: Let \( a \) and \( d \) be the first term and common difference respectively of the given A.P. Then,
\( \frac{1}{n} = m^{th} \text{ term} \Rightarrow \frac{1}{n} = a + (m – 1) d \dots (i) \)
\( \frac{1}{m} = n^{th} \text{ term} \Rightarrow \frac{1}{m} = a + (n – 1) d \dots (ii) \)
On subtracting equation (ii) from equation (i), we get
\( \frac{1}{n} – \frac{1}{m} = (m – n) d \)
\( \Rightarrow \frac{m - n}{mn} = (m – n) d \Rightarrow d = \frac{1}{mn} \)
Putting \( d = \frac{1}{mn} \) in equation (i), we get
\( \frac{1}{n} = a + \frac{(m - 1)}{mn} \Rightarrow a = \frac{1}{mn} \)
\( \therefore (mn)^{th} \text{ term} = a + (mn – 1) d = \frac{1}{mn} + (mn – 1) \frac{1}{mn} = 1 \)

Question. If \( m \) times \( m^{th} \) term of an A.P. is equal to \( n \) times its \( n^{th} \) term, show that the \( (m + n) \) term of the A.P. is zero.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then, \( m \) times \( m^{th} \) term \( = n \) times \( n^{th} \) term
\( \Rightarrow m a_m = n a_n \)
\( \Rightarrow m\{a + (m – 1) d\} = n \{a + (n – 1) d\} \)
\( \Rightarrow m\{a + (m – 1) d\} – n\{a + (n – 1) d\} = 0 \)
\( \Rightarrow a(m – n) + \{m (m – 1) – n(n – 1)\} d = 0 \)
\( \Rightarrow a(m – n) + (m – n) (m + n – 1) d = 0 \)
\( \Rightarrow (m – n) \{a + (m + n – 1) d\} = 0 \)
\( \Rightarrow a + (m + n – 1) d = 0 \)
\( \Rightarrow a_{m+n} = 0 \)
Hence, the \( (m + n)^{th} \) term of the given A.P. is zero.

Question. If the \( p^{th} \) term of an A.P. is \( q \) and the \( q^{th} \) term is \( p \), prove that its \( n^{th} \) term is \( (p + q – n) \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then,
\( p^{th} \text{ term} = q \Rightarrow a + (p – 1) d = q \dots (i) \)
\( q^{th} \text{ term} = p \Rightarrow a + (q – 1) d = p \dots (ii) \)
Subtracting equation (ii) from equation (i), we get \( (p – q) d = (q – p) \Rightarrow d = – 1 \)
Putting \( d = – 1 \) in equation (i), we get \( a = (p + q – 1) \)
\( n^{th} \text{ term} = a + (n – 1) d = (p + q – 1) + (n – 1) \times (–1) = (p + q – n) \)

If \( p^{th}, q^{th} \) and \( r^{th} \) terms of an A.P. are \( a, b, c \) respectively, then show that

Question. \( a (q – r) + b(r – p) + c(p – q) = 0 \)
Answer: Let \( A \) be the first term and \( D \) be the common difference of the given A.P. Then,
\( a = p^{th} \text{ term} \Rightarrow a = A + (p – 1) D \dots (i) \)
\( b = q^{th} \text{ term} \Rightarrow b = A + (q – 1) D \dots (ii) \)
\( c = r^{th} \text{ term} \Rightarrow c = A+ (r – 1) D \dots (iii) \)
We have, \( a(q – r) + b (r – p) + c (p – q) \)
\( = \{A + (p – 1) D\} (q – r) + \{A + (q – 1) D\} (r – p) + \{A + (r – 1) D\} (p – q) \)
\( = A \{(q – r) + (r – p) + (p – q)\} + D \{(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)\} \)
\( = A \cdot 0 + D \{p (q – r) + q (r – p) + r (p – q) – (q – r) – (r – p) – (p – q)\} \)
\( = A \cdot 0 + D \cdot 0 = 0 \)

Question. \( (a – b) r + (b – c) p + (c – a) q = 0 \)
Answer: On subtracting equation (ii) from equation (i), equation (iii) from equation (ii) and equation (i) from equation (iii), we get
\( a – b = (p – q) D, (b – c) = (q – r) D \) and \( c – a = (r – p) D \)
\( \therefore (a – b) r + (b – c) p + (c – a) q \)
\( = (p – q) Dr + (q – r) Dp + (r – p) Dq \)
\( = D \{(p – q) r + (q – r) p + (r – p) q\} = D \times 0 = 0 \)

Question. Determine the 10th term from the end of the A.P. \( 4, 9, 14, ........, 254 \).
Answer: We have, \( l = \) Last term \( = 254 \) and, \( d = \) Common difference \( = 5 \),
10th term from the end \( = l – (10 – 1) d = l – 9d = 254 – 9 \times 5 = 209 \).

ARITHMETIC MEAN (A.M.)

If three or more than three terms are in A.P., then the numbers lying between first and last term are known as Arithmetic Means between them.i.e. The A.M. between the two given quantities \( a \) and \( b \) is \( A \) so that \( a, A, b \) are in A.P.
i.e. \( A – a = b – A \Rightarrow A = \frac{a + b}{2} \)

Note : A.M. of any \( n \) positive numbers \( a_1, a_2, \dots, a_n \) is \( A = \frac{a_1 + a_2 + a_3 + \dots + a_n}{n} \)

\( n \) AM’s between two given numbers: If in between two numbers ‘a’ and ‘b’ we have to insert \( n \) AM \( A_1, A_2, \dots, A_n \) then \( a, A_1, A_2, A_3 \dots A_n, b \) will be in A.P. The series consist of \( (n + 2) \) terms and the last term is \( b \) and first term is \( a \).
\( a + (n + 2 – 1) d = b \Rightarrow d = \frac{b - a}{n + 1} \)
\( A_1 = a + d, A_2 = a + 2d, \dots A_n = a + nd \) or \( A_n = b – d \)

Note :
(i) Sum of \( n \) AM’s inserted between \( a \) and \( b \) is equal to \( n \) times the single AM between \( a \) and \( b \) i.e. \( \sum_{r=1}^{n} A_r = nA \) where \( A = \frac{a + b}{2} \)
(ii) between two numbers \( = \frac{\text{sum of m AM's}}{\text{sum of n AM's}} = \frac{m}{n} \)

SUPPOSITION OF TERMS IN A.P.

• When no. of terms be odd then we take three terms are as: \( a – d, a, a + d \). Five terms are as \( a – 2d, a – d, a, a + d, a + 2d \). Here we take middle term as ‘a’ and common difference as ‘d’.
• When no. of terms be even then we take 4 term are as : \( a – 3d, a – d, a + d, a + 3d \). 6 term are as \( a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d \). Here we take \( ‘a – d, a + d’ \) as middle terms and common difference as \( ‘2d’ \).

Note :
(i) If no. of terms in any series is odd then only one middle term is exist which is \( (\frac{n+1}{2})^{th} \) term where \( n \) is odd.
(ii) If no. of terms in any series is even then middle terms are two which are given by \( (n/2)^{th} \) and \( (\frac{n}{2} + 1)^{th} \) term where \( n \) is even.

SOME PROPERTIES OF AN A.P.

• If each term of a given A.P. be increased, decreased, multiplied or divided by some non zero constant number then resulting series thus obtained will also be in A.P.
• In an A.P., the sum of terms equidistant from the beginning and end is constant and equal to the sum of first and last term.
• Any term of an AP (except the first term) is equal to the half of the sum of terms equidistant from the term i.e. \( a_n = \frac{1}{2}(a_{n–k} + a_{n+k}), k < n \).
• If in a finite AP, the number of terms be odd, then its middle term is the AM between the first and last term and its sum is equal to the product of middle term and no. of terms.

SOME STANDARD RESULTS

• Sum of first \( n \) natural numbers \( \Rightarrow \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \)
• Sum of first \( n \) odd natural numbers \( \Rightarrow \sum_{r=1}^{n} (2r - 1) = n^2 \)
• Sum of first \( n \) even natural numbers \( = \sum_{r=1}^{n} 2r = n ( n+ 1) \)
• Sum of squares of first \( n \) natural numbers \( = \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \)
• Sum of cubes of first \( n \) natural numbers \( = \sum_{r=1}^{n} r^3 = \left[ \frac{n(n + 1)}{2} \right]^2 \)
• If for an A.P. \( p^{th} \) term is \( q, q^{th} \) term is \( p \) then \( m^{th} \) term is \( = p + q – m \)
• If for an AP sum of \( p \) terms is \( q, sum of q \) terms is \( p, \) then sum of \( (p+q) \) term is \( -(p+q). \)
• If for an A.P. sum of \( p \) terms is equal to sum of \( q \) terms then sum of \( (p + q) \) terms is zero.

EXAMPLES

Question. The sum of three numbers in A.P. is –3, and their product is 8. Find the numbers.
Answer: Let the numbers be \( (a – d), a, (a + d) \). Then,
Sum \( = – 3 \Rightarrow (a – d) + a + (a + d) = – 3 \Rightarrow 3a = – 3 \Rightarrow a = – 1 \)
Product \( = 8 \Rightarrow (a – d) (a) (a + d) = 8 \Rightarrow a (a^2 – d^2) = 8 \Rightarrow (–1) (1 – d^2) = 8 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3 \)
If \( d = 3 \), the numbers are \( –4, –1, 2 \). If \( d = – 3 \), the numbers are \( 2, – 1, –4 \). Thus, the numbers are \( –4, –1, 2 \), or \( 2, – 1, – 4 \).

Question. Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Answer: Let the numbers be \( (a – 3d), (a – d), (a + d), (a + 3d) \), Then
Sum \( = 20 \Rightarrow (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20 \Rightarrow 4a = 20 \Rightarrow a = 5 \)
Sum of the squares \( = 120 \Rightarrow (a – 3d)^2 + (a – d)^2 + (a + d)^2 + (a + 3d)^2 = 120 \)
\( \Rightarrow 4a^2 + 20d^2 = 120 \Rightarrow a^2 + 5d^2 = 30 \Rightarrow 25 + 5d^2 = 30 \) [\( \because a = 5 \)]
\( \Rightarrow 5d^2 = 5 \Rightarrow d = \pm 1 \)
If \( d = 1 \), then the numbers are \( 2, 4, 6, 8 \). If \( d = – 1 \), then the numbers are \( 8, 6, 4, 2 \). Thus, the numbers are \( 2, 4, 6, 8 \) or \( 8, 6, 4, 2 \).

Question. Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7 : 15.
Answer: Let the four parts be \( (a – 3d), (a – d), (a + d) \) and \( (a + 3d) \). Then,
Sum \( = 32 \Rightarrow (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32 \Rightarrow 4a = 32 \Rightarrow a = 8 \)
It is given that \( \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15} \)
\( \Rightarrow \frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15} \Rightarrow \frac{64 - 9d^2}{64 - d^2} = \frac{7}{15} \)
\( \Rightarrow 128d^2 = 512 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \)
Thus, the four parts are \( a – 3d, a – d, a + d \text{ and } a + 3d \) i.e. \( 2, 6, 10 \) and \( 14 \).

Question. Find the sum of 20 terms of the A.P. \( 1, 4, 7, 10, ...... \)
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then, we have \( a = 1 \) and \( d = 3 \). We have to find the sum of 20 terms of the given A.P.
Putting \( a = 1, d = 3, n = 20 \) in \( S_n = \frac{n}{2} [2a + (n – 1) d] \), we get
\( S_{20} = \frac{20}{2} [2 \times 1 + (20 – 1) \times 3] = 10 \times 59 = 590 \)

Question. Find the sum of first 30 terms of an A.P. whose second term is 2 and seventh term is 22.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then, \( a_2 = 2 \) and \( a_7 = 22 \Rightarrow a + d = 2 \) and \( a + 6d = 22 \). Solving these two equations, we get \( a = – 2 \) and \( d = 4 \).
\( S_n = \frac{n}{2} [2a + (n – 1) d] \Rightarrow S_{30} = \frac{30}{2} [2 \times (–2) + (30 – 1) \times 4] \)
\( \Rightarrow 15 (–4 + 116) = 15 \times 112 = 1680 \). Hence, the sum of first 30 terms is 1680.

Question. Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3.
Answer: Clearly, the numbers between 250 and 1000 which are divisible by 3 are \( 252, 255, 258, \dots, 999 \). This is an A.P. with first term \( a = 252 \), common difference \( = 3 \) and last term \( = 999 \). Let there be \( n \) terms in this A.P. Then,
\( \Rightarrow a_n = 999 \Rightarrow a + (n – 1)d = 999 \Rightarrow 252 + (n – 1) \times 3 = 999 \Rightarrow n = 250 \)
\( \therefore \) Required sum \( = S_n = \frac{n}{2} [a + l] = \frac{250}{2} [252 + 999] = 156375 \)

Question. How many terms of the series \( 54, 51, 48, .... \) be taken so that their sum is \( 513 \)? Explain the double answer.
Answer: \( \because a = 54, d = – 3 \text{ and } S_n = 513 \)
\( \Rightarrow \frac{n}{2} [2a + (n – 1) d] = 513 \Rightarrow \frac{n}{2} [108 + (n – 1) \times – 3] = 513 \)
\( \Rightarrow n^2 – 37n + 342 = 0 \Rightarrow (n – 18) (n – 19) = 0 \Rightarrow n = 18 \text{ or } 19 \)
Here, the common difference is negative, So, 19th term is \( a_{19} = 54 + (19 – 1) \times – 3 = 0 \). Thus, the sum of 18 terms as well as that of 19 terms is 513.

Question. If the mth term of an A.P. is \( 1/n \) and the nth term is \( 1/m \), show that the sum of \( mn \) terms is \( \frac{1}{2}(mn + 1) \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then,
\( a_m = \frac{1}{n} \Rightarrow a + (m – 1) d = \frac{1}{n} \dots (i) \) and \( a_n = \frac{1}{m} \Rightarrow a + (n – 1) d = \frac{1}{m} \dots (ii) \)
Subtracting equation (ii) from equation (i), we get \( (m – n) d = \frac{1}{n} – \frac{1}{m} \Rightarrow (m – n) d = \frac{m - n}{mn} \Rightarrow d = \frac{1}{mn} \)
Putting \( d = \frac{1}{mn} \) in equation (i), we get \( a + (m – 1) \frac{1}{mn} = \frac{1}{n} \Rightarrow a + \frac{1}{n} – \frac{1}{mn} = \frac{1}{n} \Rightarrow a = \frac{1}{mn} \)
Now, \( S_{mn} = \frac{mn}{2} \{2a + (mn – 1) \times d\} \Rightarrow S_{mn} = \frac{mn}{2} [\frac{2}{mn} + (mn - 1) \frac{1}{mn}] = \frac{1}{2} (mn + 1) \)

Question. If the sum of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its \( (m + n) \) terms is zero.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then, \( S_m = S_n \)
\( \Rightarrow \frac{m}{2} [2a + (m – 1) d] = \frac{n}{2} [2a + (n – 1) d] \)
\( \Rightarrow 2a(m – n) + \{m (m – 1) – n (n – 1)\} d = 0 \)
\( \Rightarrow 2a (m – n) + \{(m^2 – n^2) – (m – n)\} d = 0 \)
\( \Rightarrow (m – n) [2a + (m + n – 1) d] = 0 \Rightarrow 2a + (m + n – 1) d = 0 \dots (i) \) [\( \because m – n \neq 0 \)]
Now, \( S_{m+n} = \frac{m + n}{2} \{2a + (m + n – 1) d\} \Rightarrow S_{m+n} = \frac{m + n}{2} \times 0 = 0 \) [Using equation (i)]

Question. The sum of \( n, 2n, 3n \) terms of an A.P. are \( S_1, S_2, S_3 \) respectively. Prove that \( S_3 = 3(S_2 – S_1) \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then,
\( S_1 = \frac{n}{2} \{2a + (n – 1)d\} \dots (i) \)
\( S_2 = \frac{2n}{2} [2a + (2n – 1) d] \dots (ii) \)
and, \( S_3 = \frac{3n}{2} [2a + (3n – 1) d] \dots (iii) \)
Now, \( S_2 – S_1 = \frac{2n}{2} [2a + (2n – 1) d] – \frac{n}{2} [2a + (n – 1) d] = \frac{n}{2} [2 \{2a + (2n – 1)d\} – \{2a + (n – 1)d\}] = \frac{n}{2} [2a + (3n – 1) d] \)
\( \therefore 3(S_2 – S_1) = \frac{3n}{2} [2a + (3n – 1) d] = S_3 \) [Using (iii)]

Question. The sum of \( n \) terms of three arithmetical progression are \( S_1, S_2 \text{ and } S_3 \). The first term of each is unity and the common differences are \( 1, 2 \text{ and } 3 \) respectively. Prove that \( S_1 + S_3 = 2S_2 \).
Answer: We have, \( S_1 = \) Sum of \( n \) terms of an A.P. with first term \( 1 \) and common difference \( 1 = \frac{n}{2} [2 \times 1 + (n – 1) 1] = \frac{n}{2} [n + 1] \)
\( S_2 = \) Sum of \( n \) terms of an A.P. with first term \( 1 \) and common difference \( 2 = \frac{n}{2} [2 \times 1 + (n – 1) \times 2] = n^2 \)
\( S_3 = \) Sum of \( n \) terms of an A.P. with first term \( 1 \) and common difference \( 3 = \frac{n}{2} [2 \times 1 + (n – 1) \times 3] = \frac{n}{2} (3n – 1) \)
Now, \( S_1 + S_3 = \frac{n}{2} (n + 1) + \frac{n}{2} (3n – 1) = 2n^2 \text{ and } S_2 = n^2 \). Hence \( S_1 + S_3 = 2S_2 \)

Question. The sum of the first \( p, q, r \) terms of an A.P. are \( a, b, c \) respectively. Show 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 given A.P. Then, \( a = \text{Sum of } p \text{ terms} \Rightarrow a = \frac{p}{2} [2A + (p – 1) D] \Rightarrow \frac{2a}{p} = [2A + (p – 1) D] \dots (i) \)
Similarly, \( \frac{2b}{q} = [2A + (q – 1) D] \dots (ii) \) and \( \frac{2c}{r} = [2A + (r – 1) D] \dots (iii) \)
Multiplying equations (i), (ii) and (iii) by \( (q – r), (r – p) \text{ and } (p – q) \) respectively and adding, we get
\( \frac{2a}{p} (q – r) + \frac{2b}{q} (r – p) + \frac{2c}{r} (p – q) = [2A + (p – 1) D] (q – r) + [2A + (q – 1) D] (r – p) + [2A + (r – 1) D] (p – q) \)
\( = 2A (q – r + r – p + p – q) + D [(p – 1) (q – r) + (q – 1)(r – p) + (r – 1) (p – q)] \)
\( = 2A \times 0 + D \times 0 = 0 \)

Question. The ratio of the sum of \( n \) terms of two A.P.’s is \( (7n + 1) : (4n + 27) \). Find the ratio of their \( m^{th} \) terms.
Answer: Let \( a_1, a_2 \) be the first terms and \( d_1, d_2 \) the common differences of the two given A.P.’s. Then the sums of their \( n \) terms are given by \( S_n = \frac{n}{2} [2a_1 + (n – 1) d_1] \), and \( S_n' = \frac{n}{2} [2a_2 + (n – 1) d_2] \).
\( \therefore \frac{S_n}{S_n'} = \frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{7n + 1}{4n + 27} \dots (i) \)
To find the ratio of the \( m^{th} \) terms, we replace \( n \) by \( (2m – 1) \) in equation (i). Then we get
\( \frac{2a_1 + (2m - 2)d_1}{2a_2 + (2m - 2)d_2} = \frac{7(2m - 1) + 1}{4(2m - 1) + 27} \Rightarrow \frac{a_1 + (m - 1)d_1}{a_2 + (m - 1)d_2} = \frac{14m - 6}{8m + 23} \). Hence the ratio is \( (14m – 6) : (8m + 23) \).

Question. The ratio of the sums of \( m \) and \( n \) terms of an A.P. is \( m^2 : n^2 \). Show that the ratio of the \( m^{th} \) and \( n^{th} \) terms is \( (2m – 1) : (2n – 1) \).
Answer: Let \( a \) be the first term and \( d \) the common difference of the given A.P. Then, \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \Rightarrow \frac{m/2 [2a + (m - 1)d]}{n/2 [2a + (n - 1)d]} = \frac{m^2}{n^2} \Rightarrow \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \)
\( \Rightarrow [2a + (m – 1) d] n = \{2a + (n – 1) d\} m \Rightarrow 2a (n – m) = d \{(n – 1) m – (m – 1) n\} \Rightarrow 2a (n – m) = d (n – m) \Rightarrow d = 2a \)
Now, \( \frac{T_m}{T_n} = \frac{a + (m - 1)d}{a + (n - 1)d} = \frac{a + (m - 1)2a}{a + (n - 1)2a} = \frac{2m - 1}{2n - 1} \)

Question. If 4 AM’s are inserted between \( 1/2 \) and \( 3 \) then find 3rd AM.
Answer: Here \( d = \frac{3 - 1/2}{4 + 1} = \frac{1}{2} \). \( \therefore A_3 = a + 3d \Rightarrow \frac{1}{2} + 3 \times \frac{1}{2} = 2 \)

Question. \( n \) AM’s are inserted between \( 2 \) and \( 38 \). If third AM is \( 14 \) then \( n \) is equal to.
Answer: Here \( 2 + 3d = 14 \Rightarrow d = 4 \). \( \therefore 4 = \frac{38 - 2}{n + 1} \Rightarrow 4n + 4 = 36 \Rightarrow n = 8 \)

Question. Four numbers are in A.P. If their sum is 20 and the sum of their square is 120, then find the middle terms.
Answer: Let the numbers be \( a – 3d, a – d, a + d, a + 3d \). Given \( a – 3d + a – d + a + d + a + 3d = 20 \Rightarrow 4a = 20 \Rightarrow a = 5 \)
and \( (a – 3d)^2 + (a – d)^2 + (a + d)^2 + (a + 3d)^2 = 120 \Rightarrow 4a^2 + 20 d^2 = 120 \Rightarrow 4 \times 5^2 + 20 d^2 = 120 \Rightarrow d^2 = 1 \Rightarrow d = \pm 1 \). Hence numbers are \( 2, 4, 6, 8 \).

Question. Find the common difference of an AP, whose first term is \( 5 \) and the sum of its first four terms is half the sum of the next four terms.
Answer: ATQ, \( a_1 + a_2 + a_3 + a_4 = \frac{1}{2}(a_5 + a_6 + a_7 + a_8) \Rightarrow 2[a_1 + a_2 + a_3 + a_4] = a_5 + a_6 + a_7 + a_8 \)
Adding \( (a_1 + a_2 + a_3 + a_4) \) to both sides: \( 3(a_1 + a_2 + a_3 + a_4) = a_1 + \dots + a_8 \Rightarrow 3S_4 = S_8 \)
\( \Rightarrow 3 \left[ \frac{4}{2} (2 \times 5 + (4 - 1) d) \right] = \left[ \frac{8}{2} (2 \times 5 + (8 - 1) d) \right] \Rightarrow 3[10 + 3d] = 2[10 + 7d] \Rightarrow 30 + 9d = 20 + 14d \Rightarrow 5d = 10 \Rightarrow d = 2 \)

Question. If the nth term of an AP is \( (2n + 1) \) then find the sum of its first three terms.
Answer: \( \because a_n = 2n + 1 \). \( a_1 = 2(1) + 1 = 3, a_2 = 2(2) + 1 = 5, a_3 = 2(3) + 1 = 7 \). \( \therefore a_1 + a_2 + a_3 = 3 + 5 + 7 = 15 \)

Question. Which term of the sequence \( 20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4} \dots \) is the first negative terms ?
Answer: The given sequence is an A.P. in which first term \( a = 20 \) and common difference \( d = –3/4 \). Let \( a_n \) is the first negative term, then \( a_n < 0 \)
\( \Rightarrow a + (n – 1) d < 0 \Rightarrow 20 + (n – 1) (–3/4) < 0 \Rightarrow 20 < (n – 1) \frac{3}{4} \Rightarrow 80 < 3 (n – 1) \Rightarrow 80 < 3n – 3 \Rightarrow 83 < 3n \Rightarrow n > \frac{83}{3} \text{ or } n > 27 \frac{2}{3} \). \( \because 28 \) is the natural number just greater than \( 27 \frac{2}{3} \). \( \therefore n = 28 \) Ans.

IMPORTANT POINTS TO BE REMEMBERED

• A succession of numbers formed and arranged according to some definite law is called a sequence. Eg: (a) 3, 7, 11, 15 (b) 2, 4, 8, 16.
• Each number of the sequence is called a term of the sequence. A sequence is said to be finite or infinite according as the number of terms in it is finite or infinite.
• If the terms of a sequence are connected by the sign of addition (+), we get a series. Eg: 3 + 7 + 11 + 15 + .....
• If the terms of a series constantly increase or decrease in numerical value, the series is called a progression.
• A series is said to be in A.P. if the difference of each term after the first term and the proceeding term is constant. The constant difference is called common difference. Eg: 1 + 3 + 5 + 7 + 9 + ........ is an A.P. with common difference 2.
• General form of an A.P. is \( a + (n – 1)d = a_n \)
• Sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2} [2a + (n – 1)d] = \frac{n}{2} (a + a_n) \)
• \( n^{th} \) term (\( a_n \)) \( = \) sum of \( n \) terms – sum of \( (n – 1) \) terms of same AP i.e. \( a_n = S_n – S_{n–1} \)
• The \( n^{th} \) term is linear in ‘n’ and \( d = \) coefficient of \( n \).
• The sum of \( n \) terms is quadratic in ‘n’ and \( d = \) double of coefficient of \( n \).
• \( S_1 = a = \) (first term of A.P.), \( S_2 = \) sum of first two terms.
• Sum of infinite terms \( = \begin{cases} \infty & \text{if } d > 0 \\ -\infty & \text{if } d < 0 \end{cases} \)