If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256 , then find the sum of first 10 terms.

Explanation

Let a d be the first term and common difference, respectively of an AP.
`:' ` Sum of n terms of an AP, `S_(n)=(n)/(2)[2a+(n-1)d] " " `…(i)
Now, `S_(6)=36 " " ` [given]
`implies(6)/(2)[2a+(6-1)d]=36`
`implies 2a+5d=12 " " `...(i)
and `S_(16)=256`
`implies (16)/(2)[2a+(16-1)d]=256`
`implies 2a+15d=32 " " `...(ii)
On subtracting Eq. (ii) from Eq. (iii), we get
`10d=20impliesd=2`
From Eq. (ii), `2a+5(2)=12`
`implies 2a=12-10=2`
`impliesa=1`
` :. S_(10)=(10)/(2)[2a+(10-1)d]`
` " " =5[2(1)+9(2)]=5(2+18)`
` " " =5xx20=100`
Hence, the required sum of first 10 terms is 100.