Q. 204.6( 13 Votes )

In an AP, it is given that, S5 + S7 =167 and S10= 235, then find the AP, where Sn denotes the sum of its first n terms.

Answer :

Let the first term be a.

Let Common difference be d.


Given: S5 + S7 =167


S10= 235


Now, Sum of n terms of an arithmetic series is given by:


Sn = [2a + (n - 1)d]


So, consider


S5 + S7 =167


(5/2) [2a + (5 - 1)d] + (7/2) [2a + (7 - 1)d] = 167


(5/2) [2a + 4d] + (7/2) [2a + 6d] = 167


5 × [a + 2d] + 7 × [a + 3d] = 167


5a + 10d + 7a + 21d = 167


12a + 31d = 167 …………….. (1)


Now, consider S10= 235


(10/2) [2a + (10 - 1)d] = 235


5 × [2a + 9d] = 235


10a + 45d = 235


2a + 9d = 47 ………. (2)


Subtracting equation (1) from 6 times of equation (2), we get,


23d = 115


d = 5


So, from equation (2), we get,


a = 1/2 (47 - 9d)


a = 1/2 (47 - 45)


a = 1/2 (2)


a = 1


Therefore the AP is a, a + d, a + 2d, a + 3d,…


i.e. 1, 6, 11, 16, ….


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