Samacheer Kalvi Class 11 Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Exercise 5.3

 

Question 1. Find the sum of the first 20 terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.
Answer: The formula for the sum of the first \( n \) terms of an arithmetic progression (AP) is \( S_n = \frac{n}{2} [2a + (n - 1)d] \), where \( a \) is the first term and \( d \) is the common difference.
Given that the sum of the first 10 terms, \( S_{10} \), is 52.
So, \( 52 = \frac{10}{2} [2a + (10 - 1)d] \)
\( 52 = 5 [2a + 9d] \)
\( 52 = 10a + 45d \) ......(1)
Also, given that the sum of the first 15 terms, \( S_{15} \), is 77.
So, \( 77 = \frac{15}{2} [2a + (15 - 1)d] \)
\( 77 = \frac{15}{2} [2a + 14d] \)
\( 77 = 15 [a + 7d] \)
\( 77 = 15a + 105d \) ......(2)
To find \( a \) and \( d \), we solve the system of equations (1) and (2).
Multiply equation (1) by 15:
\( 52 \times 15 = 15 \times (10a + 45d) \)
\( 780 = 150a + 675d \) ......(3)
Multiply equation (2) by 10:
\( 77 \times 10 = 10 \times (15a + 105d) \)
\( 770 = 150a + 1050d \) ......(4)
Subtract equation (3) from equation (4):
\( (150a + 1050d) - (150a + 675d) = 770 - 780 \)
\( 375d = -10 \)
\( d = \frac{-10}{375} = \frac{-2}{75} \)
Now substitute the value of \( d \) into equation (1):
\( 52 = 10a + 45 \left(\frac{-2}{75}\right) \)
\( 52 = 10a - \frac{90}{75} \)
\( 52 = 10a - \frac{6}{5} \)
\( 10a = 52 + \frac{6}{5} \)
\( 10a = \frac{260 + 6}{5} \)
\( 10a = \frac{266}{5} \)
\( a = \frac{266}{5 \times 10} = \frac{266}{50} = \frac{133}{25} \)
We need to find the sum of the first 20 terms, \( S_{20} \).
\( S_{20} = \frac{20}{2} [2a + (20 - 1)d] \)
\( S_{20} = 10 \left[2\left(\frac{133}{25}\right) + 19\left(\frac{-2}{75}\right)\right] \)
\( S_{20} = 10 \left[\frac{266}{25} - \frac{38}{75}\right] \)
To subtract the fractions, find a common denominator, which is 75.
\( S_{20} = 10 \left[\frac{266 \times 3}{25 \times 3} - \frac{38}{75}\right] \)
\( S_{20} = 10 \left[\frac{798}{75} - \frac{38}{75}\right] \)
\( S_{20} = 10 \left[\frac{798 - 38}{75}\right] \)
\( S_{20} = 10 \left[\frac{760}{75}\right] \)
Simplify the fraction: Divide 760 and 75 by 5.
\( S_{20} = 10 \left[\frac{152}{15}\right] \)
\( S_{20} = \frac{1520}{15} \)
Now, simplify \( \frac{1520}{15} \) by dividing by 5.
\( S_{20} = \frac{304}{3} \)
The sum of the first 20 terms is \( \frac{304}{3} \).
In simple words: First, we use the given sums for 10 and 15 terms to find the first term and the common difference of the pattern. Then, we use these values to calculate the total sum for the first 20 terms.

🎯 Exam Tip: When dealing with systems of equations for APs, choose a method (substitution or elimination) that minimizes calculation errors and always double-check the signs and arithmetic.

 

Question 2. Find the sum up to the \( 17^{th} \) term of the series \( \frac{1^3}{1} + \frac{1^3 + 2^3}{1+3} + \frac{1^3 + 2^3 + 3^3}{1+3+5} + \dots \)
Answer: The given series is \( \frac{1^3}{1} + \frac{1^3 + 2^3}{1+3} + \frac{1^3 + 2^3 + 3^3}{1+3+5} + \dots \)
Let's find the \( n^{th} \) term of the series, denoted as \( T_n \).
The numerator of the \( n^{th} \) term is the sum of the first \( n \) cubes: \( 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \).
The denominator of the \( n^{th} \) term is the sum of the first \( n \) odd numbers: \( 1+3+5+\dots+(2n-1) = n^2 \).
So, the \( n^{th} \) term \( T_n \) is:
\( T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{n^2} \)
\( T_n = \frac{\frac{n^2(n+1)^2}{4}}{n^2} \)
\( T_n = \frac{n^2(n+1)^2}{4n^2} \)
\( T_n = \frac{(n+1)^2}{4} \)
\( T_n = \frac{n^2 + 2n + 1}{4} \)
Now, we need to find the sum of the first \( n \) terms, \( S_n \).
\( S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{k^2 + 2k + 1}{4} \)
\( S_n = \frac{1}{4} \sum_{k=1}^{n} (k^2 + 2k + 1) \)
\( S_n = \frac{1}{4} \left[ \sum_{k=1}^{n} k^2 + 2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \right] \)
We know the formulas for the sum of squares, sum of natural numbers, and sum of ones:
\( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
\( \sum_{k=1}^{n} 1 = n \)
Substitute these formulas into the expression for \( S_n \):
\( S_n = \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} + n \right] \)
\( S_n = \frac{1}{4} \left[ \frac{n(n+1)(2n+1)}{6} + n(n+1) + n \right] \)
Factor out \( n \) from the bracket:
\( S_n = \frac{n}{4} \left[ \frac{(n+1)(2n+1)}{6} + (n+1) + 1 \right] \)
\( S_n = \frac{n}{4} \left[ \frac{2n^2 + 3n + 1}{6} + \frac{6(n+1)}{6} + \frac{6}{6} \right] \)
\( S_n = \frac{n}{4} \left[ \frac{2n^2 + 3n + 1 + 6n + 6 + 6}{6} \right] \)
\( S_n = \frac{n}{4} \left[ \frac{2n^2 + 9n + 13}{6} \right] \)
\( S_n = \frac{n}{24} (2n^2 + 9n + 13) \)
We need to find the sum to the first 17 terms, so we substitute \( n = 17 \):
\( S_{17} = \frac{17}{24} (2(17)^2 + 9(17) + 13) \)
\( S_{17} = \frac{17}{24} (2(289) + 153 + 13) \)
\( S_{17} = \frac{17}{24} (578 + 153 + 13) \)
\( S_{17} = \frac{17}{24} (744) \)
Now, divide 744 by 24: \( 744 \div 24 = 31 \).
\( S_{17} = 17 \times 31 \)
\( S_{17} = 527 \)
The sum up to the 17th term of the series is 527.
In simple words: We first found a general rule for any term in the series by looking at the patterns in the top and bottom numbers. Then, we used this rule to find a formula for the sum of many terms. Finally, we put 17 into this sum formula to get the answer.

🎯 Exam Tip: Remember the standard sum formulas for cubes, natural numbers, and odd numbers. Breaking down complex series into their \( n^{th} \) term helps simplify calculations significantly.

 

Question 3. Compute the sum of first n terms of the following series.
(i) \( 8 + 88 + 888 + 8888 + \dots \)
(ii) \( 6 + 66 + 666 + 6666 + \dots \)
Answer:
(i) Let \( S_n = 8 + 88 + 888 + 8888 + \dots \) up to \( n \) terms.
We can factor out 8:
\( S_n = 8 (1 + 11 + 111 + 1111 + \dots ) \) up to \( n \) terms.
Now, multiply and divide by 9:
\( S_n = \frac{8}{9} \times 9 (1 + 11 + 111 + 1111 + \dots ) \)
\( S_n = \frac{8}{9} (9 + 99 + 999 + 9999 + \dots ) \) up to \( n \) terms.
We can write each term inside the bracket as a difference of powers of 10:
\( 9 = 10 - 1 \)
\( 99 = 100 - 1 = 10^2 - 1 \)
\( 999 = 1000 - 1 = 10^3 - 1 \)
So, \( S_n = \frac{8}{9} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)] \)
Separate the terms into two sums:
\( S_n = \frac{8}{9} [(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ up to } n \text{ times})] \)
The first part \( (10 + 10^2 + \dots + 10^n) \) is a geometric progression (GP) with first term \( a = 10 \), common ratio \( r = 10 \), and \( n \) terms.
The sum of a GP is \( S_{GP} = \frac{a(r^n - 1)}{r - 1} \).
So, \( 10 + 10^2 + \dots + 10^n = \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9} \).
The second part \( (1 + 1 + 1 + \dots \text{ up to } n \text{ times}) = n \).
Substitute these back into the expression for \( S_n \):
\( S_n = \frac{8}{9} \left[ \frac{10(10^n - 1)}{9} - n \right] \)
\( S_n = \frac{8}{81} [10(10^n - 1) - 9n] \)
(ii) Let \( S_n = 6 + 66 + 666 + 6666 + \dots \) up to \( n \) terms.
Factor out 6:
\( S_n = 6 (1 + 11 + 111 + 1111 + \dots ) \) up to \( n \) terms.
Multiply and divide by 9:
\( S_n = \frac{6}{9} \times 9 (1 + 11 + 111 + 1111 + \dots ) \)
\( S_n = \frac{6}{9} (9 + 99 + 999 + 9999 + \dots ) \) up to \( n \) terms.
Similarly, write each term as a difference of powers of 10:
\( S_n = \frac{6}{9} [(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)] \)
\( S_n = \frac{6}{9} [(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ up to } n \text{ times})] \)
Using the sum of GP and sum of ones from part (i):
\( S_n = \frac{6}{9} \left[ \frac{10(10^n - 1)}{9} - n \right] \)
\( S_n = \frac{2}{3} \left[ \frac{10(10^n - 1)}{9} - n \right] \)
\( S_n = \frac{6}{81} [10(10^n - 1) - 9n] \)
This can also be written as \( S_n = \frac{2}{27} [10(10^n - 1) - 9n] \).
In simple words: For both parts, we first changed the repeating digit sums into a pattern of powers of 10 minus 1. Then, we used the formula for the sum of a geometric progression and a simple sum of 1s to find the total sum for \( n \) terms.

🎯 Exam Tip: These types of series problems are common. Always remember the trick of multiplying and dividing by 9 (or a similar number) to convert the terms into powers of 10 minus 1, which then allows you to use GP sum formulas.

 

Question 4. Compute the sum of first n terms of \( 1 + (1 + 4) + (1 + 4 + 4^2) + (1 + 4 + 4^2 + 4^3) + \dots \)
Answer: The given series is \( 1 + (1 + 4) + (1 + 4 + 4^2) + (1 + 4 + 4^2 + 4^3) + \dots \)
Let's find the \( n^{th} \) term of the series, denoted as \( T_n \).
The \( n^{th} \) term is \( T_n = 1 + 4 + 4^2 + 4^3 + \dots + 4^{n-1} \).
This is a geometric progression with first term \( a = 1 \), common ratio \( r = 4 \), and \( n \) terms.
The sum of a GP is \( S_{GP} = \frac{a(r^n - 1)}{r - 1} \).
So, \( T_n = \frac{1(4^n - 1)}{4 - 1} \)
\( T_n = \frac{4^n - 1}{3} \)
Now, we need to find the sum of the first \( n \) terms of the main series, \( S_n \).
\( S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{4^k - 1}{3} \)
\( S_n = \frac{1}{3} \sum_{k=1}^{n} (4^k - 1) \)
\( S_n = \frac{1}{3} \left[ \sum_{k=1}^{n} 4^k - \sum_{k=1}^{n} 1 \right] \)
The first part \( \sum_{k=1}^{n} 4^k = (4^1 + 4^2 + 4^3 + \dots + 4^n) \) is a geometric progression with first term \( a = 4 \), common ratio \( r = 4 \), and \( n \) terms.
Its sum is \( S'_{GP} = \frac{4(4^n - 1)}{4 - 1} = \frac{4(4^n - 1)}{3} \).
The second part \( \sum_{k=1}^{n} 1 = n \).
Substitute these back into the expression for \( S_n \):
\( S_n = \frac{1}{3} \left[ \frac{4(4^n - 1)}{3} - n \right] \)
\( S_n = \frac{1}{9} [4(4^n - 1) - 3n] \)
This expression gives the sum of the first \( n \) terms of the series. This type of problem often involves finding a general term that is itself a sum.
In simple words: First, we looked at each part of the main series and realized that each part is a smaller pattern (a geometric series) itself. We found a general formula for this smaller pattern. Then, we summed up all these general formulas to get the total sum for the whole series.

🎯 Exam Tip: When a series's terms are themselves sums, always find the \( n^{th} \) term first by summing the inner series. Then, sum these \( n^{th} \) terms to find the overall series sum. It helps to separate the terms into two distinct sums before applying the formulas.

 

Question 5. Find the general terms and sum to n terms of the sequence \( 1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \dots \)
Answer: The given sequence is \( 1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \dots \)
Let's examine the numerators and denominators separately.
Numerators: \( 1, 4, 7, 10, \dots \)
This is an Arithmetic Progression (AP) with first term \( a = 1 \) and common difference \( d = 4 - 1 = 3 \).
The \( n^{th} \) term of this AP is \( a_n = a + (n - 1)d = 1 + (n - 1)3 = 1 + 3n - 3 = 3n - 2 \).
Denominators: \( 1, 3, 9, 27, \dots \) (The first term is \( 1 = 3^0 \), then \( 3^1, 3^2, 3^3 \), etc.)
This is a Geometric Progression (GP) with first term \( A = 1 \) and common ratio \( R = 3 \).
The \( n^{th} \) term of this GP is \( A_n = AR^{n-1} = 1 \cdot 3^{n-1} = 3^{n-1} \).
So, the general \( n^{th} \) term of the given sequence, \( T_n \), is the \( n^{th} \) term of the numerator divided by the \( n^{th} \) term of the denominator:
\( T_n = \frac{3n - 2}{3^{n-1}} \)
This sequence is an Arithmetico-Geometric Sequence.
The sum to \( n \) terms of an Arithmetico-Geometric Sequence is given by the formula:
\( S_n = \frac{a - (a+(n-1)d)r^n}{1-r} + dr \frac{1-r^{n-1}}{(1-r)^2} \)
In this problem:
The AP part is \( 1, 4, 7, \dots \) so \( a_{AP} = 1, d_{AP} = 3 \).
The GP part is \( 1, \frac{1}{3}, \frac{1}{9}, \dots \) (if we rewrite the terms as \( 1 \cdot 1, 4 \cdot \frac{1}{3}, 7 \cdot \frac{1}{9}, \dots \))
So, for the combined sequence, the first term \( a = 1 \), common difference of AP part \( d = 3 \), and common ratio of GP part \( r = \frac{1}{3} \).
Substitute these values into the formula for \( S_n \):
\( S_n = \frac{1 - (1 + (n-1)3)\left(\frac{1}{3}\right)^n}{1 - \frac{1}{3}} + 3 \left(\frac{1}{3}\right) \frac{1 - \left(\frac{1}{3}\right)^{n-1}}{\left(1 - \frac{1}{3}\right)^2} \)
Simplify the denominator \( 1 - \frac{1}{3} = \frac{2}{3} \).
\( S_n = \frac{1 - (1 + 3n - 3)\left(\frac{1}{3^n}\right)}{\frac{2}{3}} + 1 \cdot \frac{1 - \frac{1}{3^{n-1}}}{\left(\frac{2}{3}\right)^2} \)
\( S_n = \frac{3}{2} \left[1 - \frac{3n - 2}{3^n}\right] + \frac{1 - \frac{1}{3^{n-1}}}{\frac{4}{9}} \)
\( S_n = \frac{3}{2} \left[\frac{3^n - (3n - 2)}{3^n}\right] + \frac{9}{4} \left[1 - \frac{1}{3^{n-1}}\right] \)
\( S_n = \frac{3^n - 3n + 2}{2 \cdot 3^{n-1}} + \frac{9}{4} \left[\frac{3^{n-1} - 1}{3^{n-1}}\right] \)
\( S_n = \frac{3^n - 3n + 2}{2 \cdot 3^{n-1}} + \frac{3^2}{4} \frac{3^{n-1} - 1}{3^{n-1}} \)
\( S_n = \frac{3^n - 3n + 2}{2 \cdot 3^{n-1}} + \frac{3(3^{n-1} - 1)}{4 \cdot 3^{n-2}} \) (Incorrect simplification in source at this step for \( 3^2 \), re-evaluating)
Let's re-evaluate from: \( S_n = \frac{3}{2} \left[1 - \frac{3n - 2}{3^n}\right] + \frac{9}{4} \left[1 - \frac{1}{3^{n-1}}\right] \)
\( S_n = \frac{3(3^n - 3n + 2)}{2 \cdot 3^n} + \frac{9(3^{n-1} - 1)}{4 \cdot 3^{n-1}} \)
\( S_n = \frac{3^n - 3n + 2}{2 \cdot 3^{n-1}} + \frac{9(3^{n-1} - 1)}{4 \cdot 3^{n-1}} \)
To combine, make the denominators same \( 4 \cdot 3^{n-1} \):
\( S_n = \frac{2(3^n - 3n + 2)}{4 \cdot 3^{n-1}} + \frac{9(3^{n-1} - 1)}{4 \cdot 3^{n-1}} \)
\( S_n = \frac{2 \cdot 3 \cdot 3^{n-1} - 6n + 4 + 9 \cdot 3^{n-1} - 9}{4 \cdot 3^{n-1}} \)
\( S_n = \frac{6 \cdot 3^{n-1} + 9 \cdot 3^{n-1} - 6n - 5}{4 \cdot 3^{n-1}} \)
\( S_n = \frac{15 \cdot 3^{n-1} - 6n - 5}{4 \cdot 3^{n-1}} \)
\( S_n = \frac{5 \cdot 3 \cdot 3^{n-1} - 6n - 5}{4 \cdot 3^{n-1}} \)
\( S_n = \frac{5 \cdot 3^n - 6n - 5}{4 \cdot 3^{n-1}} \)
This sum is for an Arithmetico-Geometric series. The first step for such sequences is always to find the general term by analyzing the numerator and denominator separately.
In simple words: We first broke the sequence into two simpler patterns: one for the top numbers (arithmetic) and one for the bottom numbers (geometric). Then, we combined their general rules to get the rule for any term in the main sequence. Finally, we used a special formula for this type of sequence to find the total sum for \( n \) terms.

🎯 Exam Tip: When a sequence mixes arithmetic and geometric progression (Arithmetico-Geometric), always find the general term \( T_n \) by observing the numerator and denominator patterns individually. The formula for \( S_n \) of an AGP is complex but essential to memorize.

 

Question 6. Find the value of n if the sum to n terms of the series \( \sqrt{3}+\sqrt{75}+\sqrt{243}+\dots \) is \( 435 \sqrt{3} \)
Answer: The given series is \( \sqrt{3}+\sqrt{75}+\sqrt{243}+\dots \)
Let's simplify each term to identify the pattern:
First term: \( \sqrt{3} \)
Second term: \( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \)
Third term: \( \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3} \)
So, the series is \( \sqrt{3} + 5\sqrt{3} + 9\sqrt{3} + \dots \)
This is an Arithmetic Progression (AP) with:
First term \( a = \sqrt{3} \)
Common difference \( d = 5\sqrt{3} - \sqrt{3} = 4\sqrt{3} \)
We are given that the sum to \( n \) terms, \( S_n \), is \( 435\sqrt{3} \).
The formula for the sum of \( n \) terms of an AP is \( S_n = \frac{n}{2} [2a + (n - 1)d] \).
Substitute the values:
\( 435\sqrt{3} = \frac{n}{2} [2(\sqrt{3}) + (n - 1)(4\sqrt{3})] \)
\( 435\sqrt{3} = \frac{n}{2} [2\sqrt{3} + 4n\sqrt{3} - 4\sqrt{3}] \)
\( 435\sqrt{3} = \frac{n}{2} [4n\sqrt{3} - 2\sqrt{3}] \)
Factor out \( \sqrt{3} \) from the bracket:
\( 435\sqrt{3} = \frac{n}{2} \sqrt{3} [4n - 2] \)
Divide both sides by \( \sqrt{3} \):
\( 435 = \frac{n}{2} [4n - 2] \)
Factor out 2 from the bracket:
\( 435 = \frac{n}{2} \cdot 2 [2n - 1] \)
\( 435 = n(2n - 1) \)
\( 435 = 2n^2 - n \)
Rearrange into a quadratic equation:
\( 2n^2 - n - 435 = 0 \)
We can solve this quadratic equation by factoring or using the quadratic formula.
For factoring, we look for two numbers that multiply to \( 2 \times -435 = -870 \) and add to -1. These numbers are -30 and 29.
\( 2n^2 - 30n + 29n - 435 = 0 \)
\( 2n(n - 15) + 29(n - 15) = 0 \)
\( (n - 15)(2n + 29) = 0 \)
This gives two possible values for \( n \):
\( n - 15 = 0 \implies n = 15 \)
\( 2n + 29 = 0 \implies n = -\frac{29}{2} \)
Since \( n \) represents the number of terms, it must be a positive integer.
Therefore, \( n = 15 \).
The value of \( n \) for which the sum to \( n \) terms is \( 435\sqrt{3} \) is 15. This kind of problem often appears in contests to test algebraic manipulation.
In simple words: We first simplified the terms in the series to see that it followed a simple addition pattern (an arithmetic progression). Then, we used the formula for adding up terms in such a pattern and set it equal to the given total sum. Solving the resulting equation gave us the number of terms.

🎯 Exam Tip: Always simplify terms in radical series to identify the underlying progression (AP or GP). When solving for 'n', remember that the number of terms must be a positive whole number, so discard any negative or fractional solutions.

 

Question 7. Show that the sum of \( (m+n)^{th} \) and \( (m-n)^{th} \) term of an A.P is equal to twice the \( m^{th} \) term.
Answer: Let the Arithmetic Progression (AP) be denoted by its first term \( a \) and common difference \( d \).
The \( k^{th} \) term of an AP is given by \( t_k = a + (k-1)d \).
The \( (m+n)^{th} \) term is:
\( t_{m+n} = a + ((m+n) - 1)d \)
The \( (m-n)^{th} \) term is:
\( t_{m-n} = a + ((m-n) - 1)d \)
The \( m^{th} \) term is:
\( t_m = a + (m-1)d \)
We need to show that \( t_{m+n} + t_{m-n} = 2t_m \).
Let's calculate the Left Hand Side (LHS):
LHS \( = t_{m+n} + t_{m-n} \)
LHS \( = [a + (m+n-1)d] + [a + (m-n-1)d] \)
LHS \( = a + (m+n-1)d + a + (m-n-1)d \)
Group the \( a \) terms and \( d \) terms:
LHS \( = 2a + [(m+n-1) + (m-n-1)]d \)
LHS \( = 2a + [m+n-1+m-n-1]d \)
The \( +n \) and \( -n \) terms cancel out:
LHS \( = 2a + [2m - 2]d \)
LHS \( = 2a + 2(m-1)d \)
Factor out 2:
LHS \( = 2[a + (m-1)d] \)
Now, let's look at the Right Hand Side (RHS):
RHS \( = 2t_m \)
RHS \( = 2[a + (m-1)d] \)
Since LHS \( = 2[a + (m-1)d] \) and RHS \( = 2[a + (m-1)d] \), we have:
LHS \( = \) RHS
Therefore, the sum of the \( (m+n)^{th} \) and \( (m-n)^{th} \) terms of an AP is equal to twice the \( m^{th} \) term. This demonstrates a useful property of arithmetic progressions.
In simple words: We used the general rule for finding any term in an arithmetic pattern. We wrote out the specific terms for \( m+n \), \( m-n \), and \( m \). When we added the \( (m+n)^{th} \) and \( (m-n)^{th} \) terms together, the extra parts cancelled out, leaving us with exactly twice the \( m^{th} \) term.

🎯 Exam Tip: For proofs involving AP terms, always start by defining the general \( k^{th} \) term \( t_k = a + (k-1)d \). Be careful with algebraic expansion and simplification, especially when combining terms with \( (m+n-1) \) and \( (m-n-1) \).

 

Question 8. A man repays an amount of Rs.3250 by paying Rs.20 in the first month and then increases the payment by Rs. 15 per month. How long will it take him to clear the amount?
Answer: Total amount of loan = Rs.3250.
The man's monthly payments form an Arithmetic Progression (AP).
Payment in the first month, \( a = Rs.20 \).
Increase in payment each month, common difference \( d = Rs.15 \).
We need to find the number of months, \( n \), it will take for the sum of payments to reach Rs.3250.
The sum of \( n \) terms of an AP is \( S_n = \frac{n}{2} [2a + (n - 1)d] \).
Given \( S_n = 3250 \).
Substitute the values:
\( 3250 = \frac{n}{2} [2(20) + (n - 1)(15)] \)
\( 3250 = \frac{n}{2} [40 + 15n - 15] \)
\( 3250 = \frac{n}{2} [25 + 15n] \)
Multiply both sides by 2:
\( 6500 = n(25 + 15n) \)
\( 6500 = 25n + 15n^2 \)
Rearrange into a quadratic equation:
\( 15n^2 + 25n - 6500 = 0 \)
Divide the entire equation by 5 to simplify:
\( 3n^2 + 5n - 1300 = 0 \)
We can solve this quadratic equation using factoring. We need two numbers that multiply to \( 3 \times -1300 = -3900 \) and add to 5. These numbers are 65 and -60.
\( 3n^2 + 65n - 60n - 1300 = 0 \)
Factor by grouping:
\( n(3n + 65) - 20(3n + 65) = 0 \)
\( (3n + 65)(n - 20) = 0 \)
This gives two possible values for \( n \):
\( 3n + 65 = 0 \implies n = -\frac{65}{3} \)
\( n - 20 = 0 \implies n = 20 \)
Since the number of months cannot be negative, we discard \( n = -\frac{65}{3} \).
Therefore, \( n = 20 \).
It will take the man 20 months to clear the entire loan amount. This problem shows a practical application of arithmetic progressions in finance.
In simple words: The payments made each month form a steady increasing pattern. We used the total loan amount and the pattern of payments to find out how many months it would take to pay off the entire debt. We solved this using a specific math formula for sums in such patterns.

🎯 Exam Tip: For word problems involving AP sums, carefully identify the first term \( a \), common difference \( d \), and total sum \( S_n \). Solving the resulting quadratic equation for \( n \) will yield both positive and negative solutions; always choose the positive integer value for time or number of terms.

 

Question 9. In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball, 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Answer: Number of balls = 20.
Starting point is A.
Distance of the first ball from A = 24 m.
Interval between balls = 4 m.
Distances of the balls from point A are:
Ball 1: 24 m
Ball 2: 24 + 4 = 28 m
Ball 3: 28 + 4 = 32 m
...and so on.
The contestant starts from A, runs to a ball, picks it up, and returns to A. This is done for each ball.
Distance travelled to bring the first ball: Run to 24m and back to 0m = \( 24 + 24 = 2 \times 24 = 48 \) m.
Distance travelled to bring the second ball: Run to 28m and back to 0m = \( 2 \times (24 + 4) = 2 \times 28 = 56 \) m.
Distance travelled to bring the third ball: Run to 32m and back to 0m = \( 2 \times (24 + 4 + 4) = 2 \times 32 = 64 \) m.
The sequence of distances travelled for each ball is \( 48, 56, 64, \dots \)
This is an Arithmetic Progression (AP) with:
First term \( a = 48 \)
Common difference \( d = 56 - 48 = 8 \)
We need to find the total distance run to bring back all 20 balls, which is the sum of the first 20 terms of this AP, \( S_{20} \).
The formula for the sum of \( n \) terms of an AP is \( S_n = \frac{n}{2} [2a + (n - 1)d] \).
Substitute \( n = 20, a = 48, d = 8 \):
\( S_{20} = \frac{20}{2} [2(48) + (20 - 1)(8)] \)
\( S_{20} = 10 [96 + 19 \times 8] \)
\( S_{20} = 10 [96 + 152] \)
\( S_{20} = 10 [248] \)
\( S_{20} = 2480 \) m.
The contestant would run a total of 2480 meters to bring back all the balls. This problem highlights how arithmetic progressions can describe increasing distances in a practical scenario.
In simple words: The distance the contestant runs for each ball follows a regular pattern because the balls are placed at equal spaces. We found this pattern and then used a special formula to add up all the distances run for all 20 balls to get the total distance.

🎯 Exam Tip: In such problems, visualize the movement. The key is to remember that for each ball, the contestant runs to it AND back. Formulate the distances for the first few balls to identify if it's an AP or GP before applying the sum formula.

 

Question 10. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour?
Answer: Original number of bacteria (at the beginning, hour 0) = 30.
The number of bacteria doubles every hour. This means it follows a Geometric Progression (GP) where the common ratio \( r = 2 \).
Number of bacteria at the end of 1st hour: \( 30 \times 2 = 60 \)
Number of bacteria at the end of 2nd hour: \( 60 \times 2 = 30 \times 2^2 = 120 \)
Number of bacteria at the end of 3rd hour: \( 120 \times 2 = 30 \times 2^3 = 240 \)
Number of bacteria at the end of the 2nd hour: 120
Number of bacteria at the end of the 4th hour: \( 30 \times 2^4 = 30 \times 16 = 480 \)
Number of bacteria at the end of the \( n^{th} \) hour: \( 30 \times 2^n \)
This problem demonstrates exponential growth, which is modeled by a geometric progression. Each hour, the population multiplies by a constant factor.
In simple words: The bacteria population starts at 30 and gets twice as big every hour. To find the number after a certain time, we just multiply the starting number by 2, as many times as the number of hours.

🎯 Exam Tip: "Doubles every hour" implies a geometric progression with a common ratio of 2. For such growth problems, the number after \( n \) intervals is \( \text{initial amount} \times (\text{ratio})^n \).

 

Question 11. What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays an annual interest rate of 10 % compounded annually?
Answer: This is a problem of compound interest.
Principal amount \( P = Rs.500 \)
Time period \( t = 10 \) years
Annual interest rate \( r = 10\% = 0.10 \)
The interest is compounded annually.
The formula for the amount \( A \) after \( t \) years when compounded annually is:
\( A = P(1 + r)^t \)
Substitute the given values:
\( A = 500(1 + 0.10)^{10} \)
\( A = 500(1.10)^{10} \)
To calculate \( (1.10)^{10} \), we would typically use a calculator.
\( (1.10)^{10} \approx 2.59374246 \)
\( A = 500 \times 2.59374246 \)
\( A \approx 1296.87 \)
So, Rs. 500 will amount to approximately Rs. 1296.87 in 10 years. Compound interest significantly increases the final amount compared to simple interest over longer periods.
In simple words: We put Rs. 500 in a bank that adds 10% interest each year. To find out how much money we will have after 10 years, we use a special math formula that calculates how the money grows each year based on the new total.

🎯 Exam Tip: For compound interest problems, ensure you correctly identify the principal, rate, time, and compounding frequency. Remember the formula \( A = P(1 + r/n)^{nt} \), where \( n \) is the number of times interest is compounded per year. For annual compounding, \( n=1 \).

 

Question 12. In a certain town, a viral disease caused severe health hazards, upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infected virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infections virus particles just grow over 1,50,000 units?
Answer: The number of virus particles spreads in a Geometric Progression (GP).
First term \( a = 5 \) (number of particles on the first day).
The amount of infected virus particles doubles each day, so the common ratio \( r = 2 \).
The \( n^{th} \) term of a GP is \( t_n = ar^{n-1} \).
We want to find the day \( n \) when the number of virus particles \( t_n \) just grows over 1,50,000.
So, we set up the inequality: \( t_n > 1,50,000 \)
\( ar^{n-1} > 1,50,000 \)
\( 5 \times 2^{n-1} > 1,50,000 \)
Divide by 5:
\( 2^{n-1} > \frac{1,50,000}{5} \)
\( 2^{n-1} > 30,000 \)
To solve for \( n \), take the logarithm of both sides (base 10 or natural log can be used):
\( \log(2^{n-1}) > \log(30,000) \)
Using the logarithm property \( \log(x^y) = y \log(x) \):
\( (n-1) \log(2) > \log(30,000) \)
We know \( \log(2) \approx 0.3010 \) and \( \log(30,000) = \log(3 \times 10^4) = \log(3) + \log(10^4) = \log(3) + 4 \).
\( \log(3) \approx 0.4771 \), so \( \log(30,000) \approx 0.4771 + 4 = 4.4771 \).
\( (n-1) \times 0.3010 > 4.4771 \)
\( n-1 > \frac{4.4771}{0.3010} \)
\( n-1 > 14.873 \)
\( n > 14.873 + 1 \)
\( n > 15.873 \)
Since \( n \) must be an integer (representing the day), the smallest integer day greater than 15.873 is 16.
Therefore, on the 16th day, the infectious virus particles will just grow over 1,50,000 units. This illustrates exponential growth in a real-world scenario.
In simple words: The virus particles double every day, starting from 5. We wanted to find out on which day the number of particles would go past 1,50,000. We used a math formula for things that multiply by a fixed amount repeatedly and solved it using logarithms to find the specific day.

🎯 Exam Tip: For exponential growth problems, clearly define the initial value (a), the growth factor (r), and the time period (n). Using logarithms is the standard method for solving for an exponent in an inequality, and always round up to the next whole number for "just over" or "minimum days/hours" scenarios.

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