Question

The 5th term of an A.P. is 15 and the first term is 3. What is the sum of the first 10 terms?

Answer 1

EXPLANATION.

5th term of an A.P. = 15.

First term = 3.

As we know that,

General nth term  of an A.P.

⇒ Tₙ = a + (n - 1)d.

First term = a = 3   [Given]

5th term = 15.

⇒ T₅ = a + ( 5 - 1)d. = 15

⇒ T₅ = a + 4d. = 15.

⇒ a + 4d = 15.

Put the value of a = 3 in equation, we get.

⇒ 3 + 4d = 15.

⇒ 4d = 15 - 3.

⇒ 4d = 12.

⇒ d = 3.

First term = a = 3.

Common difference = d = 3.

Sum of n terms of an A.P.

⇒ Sₙ = n/2 [2a + (n - 1)d].

Put the value in this equation, we get.

⇒ S₁₀ = 10/2 [2(3) + ( 10 - 1)3].

⇒ S₁₀ = 5[ 6 + 9(3)].

⇒ S₁₀ = 5[ 6 + 27].

⇒ S₁₀ = 5[33].

⇒ S₁₀ = 165.

                                                                                                                       

MORE INFORMATION.

Some standard results.

(1) = Sum of first n natural numbers = ∑r = n(n + 1)/2.

(2) = Sum of first n odd natural numbers = ∑(2r - 1) = n².

(3) = Sum of first n even natural numbers = ∑2r = n(n + 1).

(4) = Sum of squares of first n natural numbers = ∑r² = n(n + 1)(2n + 1)/6.

(5) = Sum of cubes of first n natural numbers = ∑r³ = [n(n + 1)/2]² = (∑r)²

Answer 2

Answer:

$\sf{S_{10}}$ = 165

Step-by-step explanation:

Question:

The 5th term of an A.P. is 15 and the first term is 3. What is the sum of the first 10 terms?

Given:

• $\sf{a_5 \: = 15}$
• a = 3

To find:

Sum of the first 10 terms ($\sf{S_{10}}$)

Solution:

$\sf{a_5 }$ = a+4d

3+4d=15

4d=12

d = $\sf\dfrac{12}{4}$

d=3

Now,

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

Note: Here, n = 10

$\sf{S_{10}}$ = $\sf\dfrac{10}{2}$[2(3)+(10-1)3]

$\implies$ $\sf{S_{10}}$ = 5[6+27]

$\implies$ $\sf{S_{10}}$ = 5 $\times$ 33

$\implies$ $\sf{S_{10}}$ = 165