The sum of the second and the fifth term of an arithmetic progression is $8$ and that of the third and the seventh term is $14$ . Find the progression.

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Solution

The $n^{t h}$ term in an $A P$ is represented as:
$a_{n} = a_{1} + (n - 1) d$
where $a 1 = 1^{s t}$ term & $d =$ common difference.
Second term, $a_{2} = a_{1} + (2 - 1) d = a_{1} + d - (i)$
fifth term, $a_{5} = a_{1} + (5 - 1) d = a_{1} + 4 d - (i i)$
Third term, $a_{3} = a_{1} + (3 - 1) d = a_{1} + 2 d - (i i i)$
seventh term, $a_{7} = a_{1} + (7 - 1) d = a_{1} + 6 d - (i v)$
$N o w, a_{2} + a_{5} = 8 \Rightarrow a_{1} + d + a_{1} + 4 d = 8 [f r o m (i) & (i i)] \Rightarrow 2 a_{1} + 5 d = 8 - (v) A n d, a_{3} + a_{7} = 14 \Rightarrow a_{1} + 2 d + a_{1} + 6 d = 14 \Rightarrow 2 a_{1} + 8 d = 14 - (v i)$
Subtracting $(v)$ from $(v i)$
$2 a_{1} + 8 d = 14 2 a_{1} + 5 d = 8 3 d = 6 \Rightarrow d = 2$
From $(v), 2 a_{1} + 5 \times 2 = 8 \Rightarrow 2 a_{1} = 8 - 10 \Rightarrow a_{1} = - 1$
$∴ T h e s e r i e s i s - 1, 1, 3, 5, 7, 9, 11.$

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