Question

If first term of an AP is 5 and the 10th term is 45, then sum of first ten terms of the AP is :

Answer 1

EXPLANATION.

• GIVEN

First term of an Ap = 5

10th term of an Ap = 45

To find Sum of ten terms of an Ap

according to the question,

Formula of nth term of an Ap

$\bigstar\blue{\boxed{\bold{a_{n} \: = a + ( n \: - 1)d}}}$

First term = a = 5

T10 = 45

a + 9d = 45

put the value of a in T10 term

5 + 9d = 45

9d = 40

d = 40/9

To find sum of ten terms

Formula of sum of nth terms

$\bigstar\orange{\boxed{\bold{s_{n} \: = \frac{n}{2}(2a \: + (n - 1)d)} }}$

$\bold{s_{10} \: = \frac{10}{2} (2(5) + 9 \times \frac{40}{9} )}$

5 ( 10+ 40 )

5 ( 50) = 250

Therefore,

sum of 10th term of an Ap = 250

Answer 2

Given ,

• First term (a) = 5
• Tenth term (a10) = 45

We know that , the nth term of an AP is given by

an = a + (n - 1)d

Thus ,

a + (10 - 1)d = 45

5 + 9d = 45

9d = 40

d = 40/9

Therefore ,

The common difference of given AP is 40/9

Now ,

The sum of first n terms of an AP is given by

$\boxed{ \sf{ S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}}$

Thus ,

$\sf \mapsto S_{10} = \frac{10}{ 2} \{ 2 \times 5 + (10 - 1) \frac{40}{9} \}5 \{10 + 40 \} \\ \\ \sf \mapsto S_{10} =5 \times 50 \\ \\ \sf \mapsto S_{10} = 250$

Therefore ,

• The sum of first ten terms of given AP is 250