The sum of first 20 terms of an A.P. is 400 and the sum of first 40 terms is 1600. Find the sum of first 10 terms.

Here we have been given the first 20 terms and 40 terms of an A.P. and we need to find the sum of first 10 terms.

We know that formula to find sum of 𝑛 terms, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Where 𝑆_n is the sum of 𝑛 terms, 𝑎 is the first term, 𝑛 is the term to be found and 𝑑 is the common difference of the A.P.

Sum of first 20 𝑡𝑒𝑟𝑚𝑠, 𝑆₂₀  can be expressed as

$\begin{array}{r}{S}_{20}=\frac{20}{2}[2a+(20-1\left)d\right]\\ \Rightarrow 400=10[2a+19d]\\ \Rightarrow 2a+19d=40\end{array}$

Sum of first 40 𝑡𝑒𝑟𝑚𝑠, 𝑆₄₀ can be expressed as

$\begin{array}{r}{S}_{40}=\frac{40}{2}[2a+(40-1\left)d\right]\\ \Rightarrow 1600=20[2a+39d]\\ \Rightarrow 2a+39d=80\end{array}$

Subtracting the two equations, 

we get, 20𝑑 = 40 

⇒ 𝑑 = 2 

Putting 𝑑 = 2 in equation (𝑖),

 we get, 2 × 𝑎 + 19 × 2 = 40 

⇒ 𝑎 = 1 

Therefore, sum of first 10 terms in the 𝐴. 𝑃 is

$S_{10}=\frac{10}{2}[2\times 1+(10-1)\times 2]$

$\begin{array}{r}\Rightarrow S_{10}=5\times 20\\ \Rightarrow S_{10}=100\end{array}$