We know that, in an A.P.,
First term = a
Common difference = d
Number of terms of an AP = n
According to the question,
We have,
S5 + S7 = 167
Using the formula for sum of n terms,
Sn = (n/2) [2a + (n-1)d]
So, we get,
(5/2) [2a + (5-1)d] + (7/2)[2a + (7-1)d] = 167
5(2a + 4d) + 7(2a + 6d) = 334
10a + 20d + 14a + 42d = 334
24a + 62d = 334
12a + 31d = 167
12a = 167 – 31d …(1)
We have,
S10 = 235
(10/2) [2a + (10-1)d] = 235
5[ 2a + 9d] = 235
2a + 9d = 47
Multiplying L.H.S and R.H.S by 6,
We get,
12a + 54d = 282
From equation (1)
167 – 31d + 54d = 282
23d = 282 – 167
23d = 115
d = 5
Substituting the value of d = 5 in equation (1)
12a = 167 – 31(5)
12a = 167 – 155
12a = 12
a = 1
We know that,
S20 = (n/2) [2a + (20 – 1)d]
= 20/(2[2(1) + 19 (5)])
= 10[ 2 + 95]
= 970
Therefore, the sum of first 20 terms is 970.