Question

In an AP a3=7and a7=15 then first term is

Answer 1

ATQ.

Solve simultaneously

We know

a3=a+2d

a7=a+6d

So...,

a+2d=7

a+6d=15

Sub.

4d=8

d=2

Now put d 'sValue for a in any of eq

a+2(2)=7

a=3 Ans...

Answer 2

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

$\green{\tt{\therefore{First\:term=3}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies a_{3}= 7 \\\\ \tt:\implies a_{7}=15 \\\\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies first\:term= ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies a_{3} = 7 \\ \\ \tt: \implies a + 2d = 7 \\ \\ \tt: \implies a = 7 - 2d - - - - - (1)\\ \\ \bold{For \: a_{7}} \\ \tt: \implies a_{7} = 15 \\ \\ \tt: \implies a + 6d = 15 \\ \\ \text{Putting \: value \: of \: a} \\ \tt: \implies 7 - 2d + 6d = 15 \\ \\ \tt: \implies 7 + 4d = 15 \\ \\ \tt: \implies 4d = 15 - 7 \\ \\ \tt: \implies 4d = 8 \\ \\ \tt: \implies d = \frac{8}{4} \\ \\ \green{\tt: \implies d = 2} \\ \\ \text{Putting \: value \: of \: d \: in \: (1)} \\\\ \tt: \implies a = 7 - 2 \times 2 \\\\ \tt: \implies a = 7 - 4 \\ \\ \green{\tt: \implies a = 3} \\ \\ \green{\tt \therefore First \: term \: is \: 3}$