Question

In an A.P. 19^th term is 52 and 38^th term is 128, find sum of first 56 terms.

Answer 1

It is given that,
a₁₉ = 52
a₃₈ = 128

Now,
$$\begin{gathered} a_n=a+\left(n-1\right)d \\ {a}_{19}=a+\left(19-1\right)d \\ \Rightarrow 52=a+18d \\ \Rightarrow a=52-18d...\left(1\right) \\ {a}_{38}=a+\left(38-1\right)d \\ \Rightarrow 128=a+37d...\left(2\right) \\ \Rightarrow 128=52-18d+37d\left(\mathrm{from}\left(1\right)\right) \\ \Rightarrow 128-52=19d \\ \Rightarrow 19d=76 \\ \Rightarrow d=4 \\ \therefore a=52-18\left(4\right) \\ =-20 \\ \mathrm{Thus},a=-20\mathrm{and}d=4. \end{gathered}$$

We know that,
$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ \therefore S_{56}=\frac{56}{2}\left(2(-20)+\left(56-1\right)\left(4\right)\right) \\ =28\left(-40+220\right) \\ =28\left(180\right) \\ =5040 \end{gathered}$$

Hence, the sum of first 56 terms is 5040.