Question

please answer this question ​

Answer 1

Sum of first 6 terms is 36.

Sn = (n/2){2a+(n-1)d}

S6 = 6/2(2a + 6-1d)

36 = 3(2a+5d)

2a+5d=12 --- (1)

Sum of first 16 terms is 256.

S16=16/2[2a+15d]

256 = 8(2a+15d)

2a+15d = 32 ------ (2)

Solve (1) and (2).

2a + 5d = 12

2a + 15d = 32

----------------------

d = 2

Place d in (1)

2a + 5d = 12

=> a = 1.

Now,

s10 =10/2(2 + (10 - 1) × d)

= 5(2 + 9 * 2)

=5 * 20

=100

Therefore, sum of first 10 terms is 100.

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