Question

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

Answer 1

$\bf{\Huge{\underline{\boxed{\rm{\red{ANSWER\::}}}}}}$

Given:

The fourth term of A.P. is 11 & the eight term exceeds twice the fourth term by 5.

To find:

The A.P. & the sum of the first 50 terms.

$\bf{\large{\underline{\sf{\green{Explanation\::}}}}}}$

We know that formula of the Arithmetic Progression:

→ an= a+(n-1)d

The fourth term of A.P. is 11.

We have,

• First term of an A.P.= a
• Common difference= d

→ a4 = a + (4-1)d

→ 11 = a+ 3d..........................(1)

Eight term exceeds twice the fourth term by 5;

→ a8= 2a4 + 5

→ a8= 2 × 11 +5

→ a8= 22+5

→ a8= 27

∴ 27= a+(8-1)d

→ 27= a+ 7d.....................(2)

Subtracting equation (1) from equation (2), we get;

→ a + 7d - a+ 3d = 27-11

→ 7d - 3d = 16

→ 4d = 16

→ d= $\cancel{\frac{16}{4} }$

→ d= 4

Putting the value of d in equation (1), we get;

→ 11= a+ 3(4)

→ 11= a+ 12

→ a= 11 -12

→ a= -1

The A.P. is -1, 3, 7, 11, 15...........

Now,

• The sum of the first 50 terms

We know that formula of the sum of an Arithmetic Progression:

→ $Sn=\frac{n}{2} [2a+(n-1)d]$

→ $S50=\cancel{\frac{50}{2}} [2(-1)+(50-1)4]$

→ $S50=25(-2+49*4)$

→ $S50=25(-2+196)$

→ $S50=25(194)$

→ S50= 4850