If the sum of first 6 terms of an A.P. is 36 and that of the first term is 256 find the sum of the first 10 terms.
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Step-by-step explanation:
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 of first 10terms = 100
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