Question

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

Answer 1

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

Answer 2

Hey there !!!!

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Sum of n terms forming an AP is :

  S=n(2a+(n-1)d)/2

n=number of terms   a=first term of series  d=common difference 

sum of first 6 terms of an A.P is 36 

n=6

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

36=3(2a+5d)

12=2a+5d-----Equation 1

Sum of first 16 terms is 256:

n=16

sum=16(2a+(16-1)d)/2

     256=8(2a+15d)

     32=2a+15d----Equation 2

Solving equation's 1 & 2

d=2

d=2 in eq 2

32=2a+15*2

32-30=2a

2=2a

a=1

Sum of 10 terms :

n=10 a=1 d=2

S= n(2a+(n-1)d)/2

   =10(2+(10-1)2)/2

   = 5(2+18)=20*5=100

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Hope this helped you.................