Question

If the third term of an A.P is 7
and the 6th term is 13 then find
the sum of first 5 terms.
*​

Answer 1

GIVEN :

third term of an A.P = 7

6th term = 13

TO FIND:

sum of first 5 terms

FORMULA USED:

Tn = a+(n-1)d

Sn = n/2[2a+(n-1)d]

WHERE => a = first term

d = common difference

n= number of terms

Tn = nth term

Sn = sum of n terms

SOLUTION :

Tn = a+(n-1)d

T3 = a+(3-1)d

7 = a+2d

7-2d=a

T6 = a+(6-1)d

13 = a+5d

13-5d = a

7-2d=13-5d

-6 = -3d

2 =d

13-5d = a

13-(5×2)=a

13-10=a

3=a

Sn = n/2[2a+(n-1)d]

S5 = 5/2[(2×3)+(5-1)2]

S5 = 5/2[6+8]

S5 = 5/2 ×14

S5 = 5 ×7

S5 = 35

ANSWER : SUM OF FIRST 5 TERMS = 35

Answer 2

        $\huge\underline{\underline{\bf\bigstar ANSWER \bigstar}}$

Given ,

Third term = 7

Sixth term = 13

▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬

$\bf\longrightarrow Common \ Difference = \dfrac{Term \ difference }{Position \ difference }$

Common Difference ,

                         $\sf d = \dfrac{13-7}{6-3}$

                            $\sf = \dfrac{6}{3}$

                            $\bf = 2$

▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬

$\sf\longrightarrow 1^{st} \ term = 3^{rd} \ term - 2d$

                 $\sf = 7 - 2\times 2$

                 $\sf = 7-3$

                 $= \bf 3$

$\sf\longrightarrow 5^{th} \ term = 1^{st} \ term + 4d$

                  $\sf = 3 + 4\times 2$

                  $\sf = 3+8$

                  $\bf = 11$

▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬

   $\sf Sum \ of \ first \ n \ terms = \dfrac{n}{2} \times (x_{n} + x_{1})$  

$\sf\longrightarrow Sum \ of \ first \ five \ terms = \dfrac{5}{2} \times (11+3)$

                                      $\sf = \dfrac{5}{2} \times 14$

                                      $= \sf 5 \times 7$

                                      $\bf = 35$

▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬ ▬▬▬

HOPE IT HELPS U ............. :)