Question

there are five terms in an ap the sum of these terms is 55 and the fourth term is 5 more than the sum of the first two terms find the terms of a p​

Answer 1

Given :

• First condition

There are five terms in an ap & the sum of these terms is 55.

• Second condition

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

To find :

• Terms of APs

Solution :

• According to the first condition

There are five terms in an ap & the sum of these terms is 55.

As we know that sum of "n" terms of AP is given as :

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

Where,

• a = first term
• d = Common difference

$\implies\sf S_{5} =\dfrac{5}{2}\[ \left [2a+ (5 - 1)d \right] = 55$

$\implies \sf \dfrac{5}{2}(2a + 4d) = 55$

$\implies \sf 2a + 4d = \cancel{55} \: \: ^{11} \times \dfrac{2}{ \cancel 5}$

$\implies \sf 2(a + 2d) = 22$

$\implies \sf a + 2d = 11 \: \: \: \bf{(equation \:1)}$

• According to the second condition

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

$\implies \sf a_4 = 5 + S_2$

$\implies \sf a + (4 - 1)d = 5 + \sf \dfrac{2}{2} \[ \left[ 2a + (2 - 1)d \right] \]$

$\implies \sf a + 3d = 5 + 2a +d$

$\implies \sf a - 2a + 3d - d = 5$

$\implies \sf - a + 2d = 5 \: \: \bf{(equation \:2)}$

• Add both the equations

→ a + 2d + (-a + 2d) = 11 + 5

→ a + 2d - a + 2d = 16

→ 4d = 16

→ d = 16/4 = 4

• Put the value of d in eqⁿ 1

→ -a + 2d = 5

→ -a + 2 × 4 = 5

→ -a + 8 = 5

→ a = 8 - 5 = 3

• Terms of AP

→ a₁ = 3

→ a₂ = a + d = 3 + 4 = 7

→ a₃ = a + 2d = 3 + 8 = 11

→ a₄ = a + 3d = 3 + 12 = 15

→ a₅ = a + 4d = 3 + 16 = 19

•°• Required A.P = 3, 7, 11, 15, 19..

Answer 2

Answer:

• 3 , 7 , 11 , 15 , 19.

Step-by-step explanation:

Given

• There are five terms in an A.P.
• The sum of these terms is 55.
• Fourth term is 5 more than the sum of first and two terms.

To find

• Terms of the A.P.

Solution

↪ Sum of all the terms in the A.P :

• 5a = 55
• a = 55/5
• a = 11

↪ Let the 5 terms be (a - 2d), (a - d), a , (a + d), (a + 2d).

ATP,

• 4rth term = 5 + (1st term + 2nd term)
• a + d = 5 + a - 2d + a - d
• 4d = 5 + a
• 4d = 5 + 11
• 4d = 16
• d = 16/4
• d = 4

Substituting in the 5 terms :

• (a - 2d) = 3
• (a - d) = 7
• (a) = 11
• (a + d) = 15
• (a + 2d) = 19

∴ The A.P is 3 , 7 , 11 , 15 , 19.