If the sum of first 6 terms of an A.P. is 36 and that of the first 16

1.	If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find thesum of first 10 terms.
Answer» Sum of n terms = n/2{2a + (n -1)d } A/C to question ; S6 = 6/2{ 2a + (6-1)d } 36 = 3(2a + 5d) 12 = 2a + 5d --------------(1) again, S16 = 16/2{ 2a + (16-1)d }256 = 8{ 2a + 15d }32 = 2a + 15d --------(2) solve eqns (1) and (2) 10d = 20 d = 2 put in equation (1) a = 1 now, sum of 10th terms = S10 = 10/2{2a +(10-1)d}= 5{ 2 × 1 + 9 ×2 }= 5 { 2 + 18}= 100 hence sum is 100

If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find thesum of first 10 terms.

Answer»

Sum of n terms = n/2{2a + (n -1)d }

A/C to question ;

S6 = 6/2{ 2a + (6-1)d } 36 = 3(2a + 5d) 12 = 2a + 5d --------------(1)

again, S16 = 16/2{ 2a + (16-1)d }256 = 8{ 2a + 15d }32 = 2a + 15d --------(2)

solve eqns (1) and (2)

10d = 20 d = 2 put in equation (1) a = 1

now, sum of 10th terms = S10 = 10/2{2a +(10-1)d}= 5{ 2 × 1 + 9 ×2 }= 5 { 2 + 18}= 100

hence sum is 100