Question

There are five terms in an arthematic progrresion the sum of these terms is 55 and the fourth term is five more than the sum of the firsrmt two terms

Answer 1

Answer :

The terms of the A.P. is 3, 7, 11, 15, 19.

Given :

• There are five terms in an arithmetic progression. The sum of these terms is 55.
• The fourth term is 5 more than the sum of the first two terms.

To find :

• Terms of the arithmetic progression.

Solution :

Consider, the AP is ,

a, a+d, a+2d, a+3d, a+4d

Formula Used :-

${\boxed{\sf{S_n=\dfrac{n}{2}[2a+(n-1)d]}}}$

• n = 5

★Sum of five terms is 55.

According to the 1st condition :-

$\to\sf{S_n=\dfrac{n}{2}[2a+(n-1)d]}$

$\to\sf{55=\dfrac{5}{2}[2a+(5-1)d]}$

$\to\sf{55=\dfrac{5}{2}[2a+4d]}$

$\to\sf{55=5(a+2d)}$

$\to\sf{a+2d=11.................(i)}$

★ The fourth term is 5 more than the sum of the first two terms.

According to the 2nd condition :-

$\to\sf{a+3d=a+a+d+5}$

$\to\sf{a+3d=2a+d+5}$

$\to\sf{2d-a=5.................(ii)}$

Now add both equations (i) & (ii) .

a+2d+2d-a=11+5

→ 4d = 16

→ d = 16/4

→ d = 4

Now put d=4 in eq(ii) .

2d - a = 5

→ 2×4 - a = 5

→ 8 - a = 5

→ - a = 5-8

→ - a = -3

→ a = 3

Then,

★ 1st term = 3

★ 2nd term = 3+4 = 7

★ 3rd term = 3 + 2×4 = 11

★ 4th term = 3 + 3×4 = 15

★ 5th term = 3 + 4×4 = 19

AP → 3,7,11,15,19

Answer 2

Given :

• There are five terms in an Arithmetic Progression. The sum of these terms is 55, and the fourth term is five more than the sum of the first two terms.

To find :

• terms of the Arithmetic progression.

Solution:

• Let The AP be

• a
• a + d
• a + 2d
• a + 3d
• a + 4d

• Sn = (n/2)[2a + (n-1)d ]
• Sum of 5 terms

=> (5/2)(2a + 4d) = 55

=> a + 2d = 11

• fourth term is five more than the sum of the first two terms.

=> a + 3d = 5 + a + a + d

=> 2d - a = 5

• Adding both

=> 4d = 16

=> d = 4

=> a = 3

$\bf{\underline{HENCE:-}}$

• A.P will be = a = 3
• a + d = 3 + 4 = 7
• a + 2d = 3 + 2×4 = 11
• a + 3d = 3 + 3 × 4 = 15
• a + 4d = 3 × 4×4 = 19

• A.P is 3 , 7, 11 , 15 , 19