Question

If the third term of an A.P is 7
and the 6th term is 18 then find
the sum of first 5 terms.​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

$A(3) = a + 2d = 7 -----(1)$

$A(6) = a + 5d = 18 -----(2)$

$From \: eq.(1) \: - eq.(2)$

$5d - 2d = 18 - 7$

$3d = 11$

$d = 11/3$

$Then , a = 7 - 2(\frac{11}{3})$

$a = 7 - \frac{22}{3}$

$a = \frac{21}{3} - \frac{22}{3}$

$a = \frac{-1}{3}$

$Now ,$

$S(5) = \frac{5}{2}[2(\frac{-1}{3}) + (5-1)(\frac{11}{3})]$

$S(5) = \frac{5}{2}[\frac{-2}{3} + 4(\frac{11}{3})]$

$S(5) = \frac{5}{2}[\frac{-2}{3} + \frac{44}{3}]$

$S(5) = \frac{5}{2} × \frac{42}{3}$

$S(5) = \frac{5}{2} × 14$

$S(5) = 5 × 7 = 35$