Question

Determine the sum of first 35 terms of an A.P., if the second term is 2 and the seventh term is 22.

Answer 1

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

$\green{\tt{\therefore{Sum\:of\:35\:term=2310}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Second\:term= 2\\\\ \tt:\implies Seventh\:term=22 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Sum \: of\:first\:35\:terms= ?$

• According to given question :

$\tt: \implies a_{2} = 2 \\ \\ \tt: \implies a + d = 2 - - - - - (1) \\ \\ \tt: \implies a_{7} = 22 \\ \\ \tt: \implies a + 6d = 22 - - - - - (2) \\ \\ \text{Subtracting \: (1) \: from \: (2)} \\ \tt: \implies 6d - d = 22 - 2 \\ \\ \tt: \implies 5d = 20 \\ \\ \green{\tt: \implies d = 4} \\ \\ \text{Putting \: value \: of \: d \: in \: (1)} \\ \tt: \implies a + 4 = 2 \\ \\ \green{\tt: \implies a = - 2} \\ \\ \tt \circ \: Number \: of \: terms = 35 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2} (2a +( n - 1)d) \\ \\ \tt: \implies s_{35} = \frac{35}{2}(2 \times - 2 + (35 - 1) \times 4 )\\ \\ \tt: \implies s_{35} = \frac{35}{2} \times ( - 4 + 136) \\ \\ \tt: \implies s_{35} = \frac{35}{2} \times 132 \\ \\ \green{\tt: \implies s_{35} = 2310}$

Answer 2

$hey$

Here

2term=a+d=2 ..................eq-1

7term=a+6d=22 ............eq-2

Substituting eq1 eq1&2

we get,

a=-2

d=4

Here n=35

Sum of the 35 terms =n/2[2a+(n-1)d]=4620