This is the basic and conceptual question from Chapter name- Arithmetic Progression

Chapter number- 9

Exercise 9.6

In this question we have been given that the third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2.

Now we have to find the first term, the common difference and the sum of first 20 terms.

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Class:- 10th

Solutions of CBSE Mathematics

Question 15

Given A.P. has third term(a

_{3}) = 7 and seventh term(a_{7}) = 3a_{3}+ 2 = 3(7) + 2 = 23.We know nth term of an A.P. is given by, a

_{n}= a + (n − 1)d. So, we get,a + 2d = 7 ….. (1)

a + 6d = 23 ….. (2)

Subtracting (1) from (2), we get,

=> (a + 6d) − (a + 2d) = 23 − 7

=> 4d = 16

=> d = 4

On putting d = 4 in (1), we get,

=> a + 2(4) = 7

=> a = 7 − 8

=> a = −1

We know sum of n terms of an A.P. is given by S

_{n}= n[2a + (n − 1)d] / 2.Here a = −1, d = −4, n = 20. So sum is,

S

_{20}= 20[2(−1) + (20 − 1)(4)]/2= 20[-2 + 76]/2

= 20[39] = 740

Hence, the sum of first 20 terms for the given A.P. is 740.