Question

In an A.P. the 10^th term is 46 sum of the 5^th and 7^th term is 52. Find the A.P.

Answer 1

It is given that,
t₁₀ = 46
t₅ + t₇ = 52

Now,
$$\begin{gathered} t_n=a+\left(n-1\right)d \\ {t}_{10}=a+\left(10-1\right)d \\ \Rightarrow 46=a+9d \\ \Rightarrow a=46-9d...\left(1\right) \\ {t}_5+t_7=52 \\ \Rightarrow \left(a+\left(5-1\right)d\right)+\left(a+\left(7-1\right)d\right)=52 \\ \Rightarrow a+4d+a+6d=52 \\ \Rightarrow 2a+10d=52 \\ \Rightarrow 2\left(46-9d\right)+10d=52\left(\mathrm{from}\left(1\right)\right) \\ \Rightarrow 92-18d+10d=52 \\ \Rightarrow 92-8d=52 \\ \Rightarrow 8d=92-52 \\ \Rightarrow 8d=40 \\ \Rightarrow d=5 \\ \Rightarrow a=46-9\left(5\right)\left(\mathrm{from}\left(1\right)\right) \\ \Rightarrow a=46-45 \\ \Rightarrow a=1 \end{gathered}$$

Hence, the given A.P. is 1, 6, 11, 16, ....