The sum of first 7 terms of an 𝐴𝑃 is 63 and the sum of its next 7 terms is161. Find the 28𝑡ℎ term of this 𝐴. 𝑃.

Given that, 

The sum of first 7 terms of an 𝐴𝑃 is 63 

The sum of its next 7 terms is 161 

Now we have to find that the 28𝑡ℎ term of this 𝐴. 𝑃 

Here, 𝑆₇ = 63 and 𝑛 = 7 The sum of terms in an 𝐴. 𝑃.

$\begin{array}{r}\Rightarrow S=\frac{n}{2}(2a+(n-1\left)d\right)\\ \Rightarrow 63=\frac{7}{2}(2a+(7-1\left)d\right)\\ \Rightarrow 63=\frac{7}{2}(2a+6d) \end{array}$

$\begin{array}{r}\Rightarrow 63=7a+21d \\ \Rightarrow 9=a+3d \\ \Rightarrow a=9-3d\dots \left(i\right) \\ \text{ Also, }{S}_{14}=63+161 \\ S_{14}=224n=14 \\ \Rightarrow 224=\frac{14}{2}(2a+(14-1\left)d\right) \\ \Rightarrow 224=7(2a+13d) \\ \Rightarrow 224=7\left(2\right(93d)\mid 13d) \\ \Rightarrow 32=18-6d+13d \\ \Rightarrow 32=18+7d \\ \Rightarrow 7d=14 \\ \Rightarrow d=2 \end{array}$

$From \left(i\right), a=9-6$

$\text{ Now, }{a}_{28}=a+27d$

$\begin{array}{r}=3+27\left(2\right) \\ =3+54 \\ =57 \\ \Rightarrow a_{28}=57 \end{array}$

Hence, the 28𝑡ℎ term of this 𝐴. 𝑃. is 57