Question

The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

Answer 1

For an A.P.

t₄ = 11

`\implies` a + 3d = 11  ...(i)

Also, t₈ – 2t₄ = 5

`\implies` (a + 7d) – 2 × 11 = 5

`\implies` a + 7d – 22 = 5

`\implies` a + 7d = 27  ...(ii)

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

a + 3d = 11
a + 7d = 27
–   –       –    
– 4d = – 16

d = `(- 16)/(- 4)`

`\implies` d = 4

Substituting d = 4 in equation (i), we get

a + 3d = 11

a + 3 × 4 = 11

`\implies` a + 12 = 11

`\implies` a = –1

∴ Required A.P. = a, a + d, a + 2d, a + 3d, ....

∴ a = –1

∴ a + d = –1 + 4 = 3

∴ a + 2d = –1 + 2(4) = –1 + 8 = 7

∴ a + 3d = –1 + 3(4) = –1 + 12 = 11

Sum of first 50 terms of this A.P.

=`50/2 [2 xx (-1) + 49 xx 4]`

= 25[–2 + 196]

= 25 × 194

= 4850