Question

Calculate the sum of 35 terms in an AP, whose fourth term is 16 and ninth term is 31.

Answer 1

Let a and d be the first term and the common difference, respectively.

Now, a₄ = 16  ......[Given]

⇒ a + 3d = 16  .....(i)

Also, a₉ = 31  ......[Given]

⇒ a + 8d = 31  .....(ii)

Subtracting the equation (i) from equation (ii), we get

a + 8d = 31
a + 3d = 16
–    –   –      
      5d  = 15

⇒ d = 3

Substituting the value of d in equation (i), w e get

a + 3 (3) = 16

⇒ a + 9 = 16

⇒ a = 16 – 9 = 7

Now, `"S"_n = n/2 [2a + (n - 1)d]`

∴ `"S"_35 = 35/2 [2(7) + (35 - 1)3]`

= `35/2 [14 + (34)3]`

= `35/2 (14 + 102)`

= `35/2 (116)`

= 35 × 58

= 2030

Hence, the sum of 35 terms of the A.P. is 2030.