Question
$51$ terms of the $A.P.$ whose second term is $2$ and fourth term is $8.$
✓
Answer
Given,
$a_2= 2$ and $a_4= 8$
$a + d = 2 .....(i)$
$a + 3d = 8 .....(ii)$
Put $d = 3 in .....(i) ⇒ a + d = 2$
$a + 3 = 2$
$a = -1$
$\text{S}_{51}=\frac{51}{2}(2\times-1+(51-1)\times3)$
$\Big(\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})\Big)$
$=\frac{51}{2}(-2+50\times3)$
$=\frac{51}{2}\times148$
$=3774$
$\therefore\ \text{S}_{\text{n}}=3774$
$a_2= 2$ and $a_4= 8$
$a + d = 2 .....(i)$
$a + 3d = 8 .....(ii)$
Put $d = 3 in .....(i) ⇒ a + d = 2$
$a + 3 = 2$
$a = -1$
$\text{S}_{51}=\frac{51}{2}(2\times-1+(51-1)\times3)$
$\Big(\therefore\ \text{S}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})\Big)$
$=\frac{51}{2}(-2+50\times3)$
$=\frac{51}{2}\times148$
$=3774$
$\therefore\ \text{S}_{\text{n}}=3774$
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