Question

Find the sum of the first 35 terms of an Arithmetic Progression whose
third term is 7 and seventh term is two more than thrice of its third
term.

Answer 1

Step-by-step explanation:

The first term, common difference and the sum of first 20 terms are -1, 4, 740 respectively.

Step-by-step explanation:

Given a3=7

A7 = 3a3 +2

A1=?

D=?

S20=?

Step 1:

A7 =3 x a3 +2

A + 6d =3x (a+ 2d) + 2

A + 6 d =3a + 6d +2

3a –a =- 2

2a = -2

A = - 1

Step 2:

A3=a+2d =7

-1 + 2d = 7

2d = 7 +1  

D = 4

Step 3:

AP= a = -1

A + d =-1 +4 =3

A +2d =-1 + 8 = 7

A.P = -1, 3, 7------ to 20 terms

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

Step 4:

S20= 20/2 [2 x -1 + (20 – 1)4]

=10 (-2 +19x 4)

= 10 (-2 x 76)

10 x 74 = 740

S20 =740.

Thanks!!!