CBSE Class 10 Mathematics Arithmetic Progression Notes Set 08

Selection of Terms and Sum of the First n Terms of an A.P.

Selection of Terms of an A.P.

Sometimes we require a certain number of terms in A.P. The following ways of selecting terms are generally very convenient.

Sum of the First n Terms of an A.P.

Let the A.P. be \( a, a + d, a + 2d, \dots, a + (n - 1)d \). Then \( S_n = a + (a + d) + \dots + \{a + (n - 2)d\} + \{a + (n - 1)d\} \) ... (i). Writing the terms in the reverse order, \( S_n = \{a + (n - 1)d\} + \{a + (n - 2)d\} + \dots + (a + d) + a \) ... (ii). Adding (i) and (ii): \( 2S_n = n[2a + (n - 1)d] \). Hence

\[ S_n = \frac{n}{2}\left[2a + (n - 1)d\right] = \frac{n}{2}\left[a + \ell\right], \]

where \( \ell = a + (n - 1)d \) is the last term.

Result. The nth term can be found from the sum: \( a_n = S_n - S_{n-1} \) (for \( n \ge 2 \)).

Solved Examples

Example 12. 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. Sum: \( 3a = -3 \Rightarrow a = -1 \). Product: \( a(a^2 - d^2) = 8 \Rightarrow -(1 - d^2) = 8 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3 \).
If a = -1 and d = 3 the numbers are -4, -1, 2. If d = -3 the numbers are 2, -1, -4. The numbers are -4, -1, 2.

Example 13. Find the sum of 20 terms of the A.P. 1, 4, 7, 10, ...
Answer:
a = 1, d = 3. \( S_{20} = \dfrac{20}{2}[2(1) + 19 \times 3] = 10 \times 59 = 590 \).

Example 14. Find the sum of all three digit natural numbers which are divisible by 7.
Answer:
The first and last three digit numbers divisible by 7 are 105 and 994. The A.P. 105, 112, ..., 994 has a = 105, d = 7. \( 994 = 105 + (n - 1)7 \Rightarrow n = 128 \).
\( S_{128} = \dfrac{128}{2}(105 + 994) = 64 \times 1099 = 70336 \).

Example 15. 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 mth and nth terms is \( (2m - 1) : (2n - 1) \).
Answer:
\( \dfrac{S_m}{S_n} = \dfrac{m^2}{n^2} \Rightarrow \dfrac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \dfrac{m^2}{n^2} \Rightarrow \dfrac{2a + (m - 1)d}{2a + (n - 1)d} = \dfrac{m}{n} \).
Cross-multiplying: \( 2an + n(m - 1)d = 2am + m(n - 1)d \Rightarrow 2a(n - m) = d(n - m) \Rightarrow d = 2a \) (as \( m \ne n \)).
\( \dfrac{a_m}{a_n} = \dfrac{a + (m - 1)2a}{a + (n - 1)2a} = \dfrac{a(2m - 1)}{a(2n - 1)} = \dfrac{2m - 1}{2n - 1} \).

Example 16. In an A.P., the sum of the first n terms is \( 4n^2 - 3n \). Find its nth term.
Answer:
\( a_n = S_n - S_{n-1} = 4n^2 - 3n - [4(n - 1)^2 - 3(n - 1)] = 4n^2 - 3n - 4n^2 + 8n - 4 + 3n - 3 = 8n - 7 \).

Check Your Progress

1. Find three numbers in A.P. whose sum is 15 and the sum of the squares of the two extremes is 58.
Answer:
Let the numbers be 5 - d, 5, 5 + d. \( (5 - d)^2 + (5 + d)^2 = 58 \Rightarrow 50 + 2d^2 = 58 \Rightarrow d = \pm 2 \). The numbers are 3, 5, 7.

2. The sum of 15 terms of an A.P. is 600 and the common difference is 5. Find the first term.
Answer:
\( 600 = \dfrac{15}{2}(2a + 70) \Rightarrow 2a + 70 = 80 \Rightarrow \) a = 5

3. Find the sum of 15 terms of the series whose nth term is 4n + 1.
Answer:
\( a_1 = 5 \), \( a_{15} = 61 \). \( S_{15} = \dfrac{15}{2}(5 + 61) = \) 495

4. If the sum of n terms of an A.P. is \( n^2 + 7n \), find the common difference.
Answer:
\( a_n = S_n - S_{n-1} = 2n + 6 \), so d = 2.

5. Find the sum of all two digit numbers which are divisible by 4.
Answer:
12, 16, ..., 96: \( 96 = 12 + (n - 1)4 \Rightarrow n = 22 \). \( S = \dfrac{22}{2}(12 + 96) = \) 1188

Word Problems

Solved Examples

Example 17. A manufacturer of TV sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production in (i) the first year (ii) the 10th year (iii) 7 years.
Answer:
The yearly production forms an A.P. with first term a and common difference d. \( a_3 = a + 2d = 600 \) and \( a_7 = a + 6d = 700 \). Solving: a = 550, d = 25.
(i) The production in the first year is 550 TV sets.
(ii) Production in the 10th year \( = a + 9d = 550 + 225 = 775 \).
(iii) Total production in 7 years \( = S_7 = \dfrac{7}{2}[2(550) + 6 \times 25] = 4375 \) TV sets.

Example 18. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find the value of x.
Answer:
The house numbers form an A.P. with first term 1 and common difference 1. We need \( S_{x-1} = S_{49} - S_x \).
\( \dfrac{x - 1}{2}[2 + (x - 2)] = \dfrac{49}{2}(1 + 49) - \dfrac{x}{2}(2 + x - 1) \Rightarrow \dfrac{x(x - 1)}{2} = 1225 - \dfrac{x(x + 1)}{2} \Rightarrow \dfrac{x}{2}[(x - 1) + (x + 1)] = 1225 \Rightarrow x^2 = 1225 \).
So x = 35 (x cannot be negative). Since x is a whole number between 1 and 49, such a house exists and its number is 35.

Example 19. A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each prize.
Answer:
Let the first prize be Rs. a. The prizes form an A.P. with d = -20 and \( S_7 = 700 \). \( \dfrac{7}{2}[2a + 6(-20)] = 700 \Rightarrow 2a - 120 = 200 \Rightarrow a = 160 \).
The prizes are Rs. 160, 140, 120, 100, 80, 60 and 40.

Check Your Progress

1. From your pocket money, you save Re. 1 on day 1, Rs. 2 on day 2, Rs. 3 on day 3, and so on. How much money will you save in the month of March 2018?
Answer:
March has 31 days: \( S_{31} = \dfrac{31}{2}(1 + 31) = \) Rs. 496

2. Shubham saves Rs. 32 during the first month, Rs. 36 in the second month and Rs. 40 in the third month. If he continues to save in this manner, in how many months will his total saving be Rs. 2000?
Answer:
a = 32, d = 4. \( \dfrac{n}{2}[64 + 4(n - 1)] = 2000 \Rightarrow n^2 + 15n - 1000 = 0 \Rightarrow (n - 25)(n + 40) = 0 \). 25 months

3. Yash started work in 2017 at an annual salary of Rs. 25000 and received a raise of Rs. 2000 each year. In what year did his income reach Rs. 45000?
Answer:
\( 25000 + (n - 1)2000 = 45000 \Rightarrow n = 11 \). The 11th year is 2017 + 10 = 2027.

4. Kanishk saves Rs. 12 in the first week of a year and then increases her weekly saving by Rs. 1.75. If in the nth week her weekly saving becomes Rs. 20.75, find n.
Answer:
\( 12 + (n - 1)(1.75) = 20.75 \Rightarrow n - 1 = 5 \Rightarrow \) n = 6