Question

Find n for the series 2 + 6 + 10 + 14 + ... if the sum of the related series is 1058

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Number\:of\:terms=23}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies A.P = 2,6,10,14,..... \\ \\ \tt: \implies Sum \: of \: n \: terms(s_{n}) = 1058 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Number \: of \: terms \: in \:A.P = ?$

• According to given question :

$\tt \circ \: First \: term = 2 \\ \\ \tt \circ \: Common \: difference = 4 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \\ \tt: \implies 1058 = \frac{n}{2} (2 \times 2 + (n - 1) \times 4) \\ \\ \tt: \implies 1058 \times 2 = n \times (4 + 4n - 4) \\ \\ \tt: \implies 1085 \times 2 = 4 {n}^{2} \\ \\ \tt: \implies 2116 = 4 {n}^{2} \\ \\ \tt: \implies {n}^{2} = \frac{2116}{4} \\ \\ \tt: \implies {n}^{2} = 529 \\ \\ \tt: \implies n = \sqrt{529} \\ \\ \green{\tt: \implies n = \pm23 }\\ \\ \green{\tt \therefore Number \: of \: terms \: in \:A.P \: is \: 23}$