Question

find the sum of first 101 terms of an AP 3,15,27,39....​

Answer 1

Answer:

,

Step-by-step explanation:

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Answer 2

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

$\green{\tt{\therefore{Sum\:of\:101\:terms=60903}}}$

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

$\green{ \underline \bold{Given: }} \\ \tt: \implies A.P = 3,15,27,39,... \\ \\ \red{ \underline \bold{To \: Find: }} \\ \tt: \implies Sum \: of \: 101 \: terms=?$

• According to given question :

$\tt \circ \: First \: term = 3 \\ \\ \tt \circ \: Common \: Difference = 12 \\ \\ \tt \circ \: Number \: of \: terms = 101 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2} (2a + (n - 1) d) \\ \\ \tt: \implies s_{101} = \frac{101}{2} (2 \times 3 + (101 - 1) \times 12) \\ \\ \tt: \implies s_{101} = \frac{101}{2} \times (6 + 100 \times 12) \\ \\ \tt: \implies s_{101} = \frac{101}{2} \times (6 + 1200) \\ \\ \tt: \implies s_{101} = \frac{101}{2} \times 1206 \\ \\ \tt: \implies s_{101} =101 \times 603 \\ \\ \green{\tt: \implies s_{101} =60903} \\ \\ \green{\tt \therefore Sum \: of \: 101 \: terms \: is \: 60903}$