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(a3) = 7 and seventh term(a7) = 3a3 + 2 = 3(7) + 2 = 23.
We know nth term of an A.P. is given by, an = 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 Sn = n[2a + (n − 1)d] / 2.
Here a = −1, d = −4, n = 20. So sum is,
S20 = 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.