# Show that a_{1}, a_{2},... , an , ... form an AP where an is defined as below :

(i) a_{n} = 3 + 4n (ii) a_{n} = 9 - 5n

Also, find the sum of the first 15 terms in each case

**Solution:**

A sequence that has common difference between any two of its consecutive terms is an arithmetic progression.

Sum of the first n terms of an AP is given by S_{n} = n/2 [2a + (n - 1) d] or S_{n} = n/2 [a + l], and the nth term of an AP is a_{n} = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.

(i) a_{n} = 3 + 4n

Given,

- nth term, a
_{n}= 3 + 4n

a_{1} = 3 + 4 × 1 = 7

a_{2} = 3 + 4 × 2 = 3 + 8 = 11

a_{3} = 3 + 4 × 3 = 3 + 12 = 15

a_{4} = 3 + 4 × 4 = 3 + 16 = 19

It can be observed that

a_{2} - a_{1} =11 - 7 = 4

a_{3} - a_{2} =15 - 11 = 4

a_{4} - a_{3} =19 - 15 = 4

So, the difference of a_{n} and a_{n - 1} is constant.

Therefore, this is an AP with common difference as 4 and first term as 7.

Sum of n terms,

S_{n} = n/2 [2a + (n - 1) d]

= 15/2 [2 × 7 + (15 - 1) 4]

= 15/2 [14 + 14 × 4]

= 15/2 × 70

= 15 × 35

= 525

(ii) a_{n} = 9 - 5n

Given,

- nth term is a
_{n}= 9 - 5n

a_{1} = 9 - 5 × 1 = 9 - 5 = 4

a_{2} = 9 - 5 × 2 = 9 - 10 = - 1

a_{3} = 9 - 5 × 3 = 9 - 15 = - 6

a_{4} = 9 - 5 × 4 = 9 - 20 = - 11

It can be observed that

a_{2} - a_{1} = (-1 ) - 4 = - 5

a_{3} - a_{2} = (-6) -(-1) = - 5

a_{4} - a_{3} = (- 11) - (- 6) = - 5

So, the difference of a_{n} and a_{n - 1} is constant.

Therefore, this is an A.P. with common difference - 5 and first term as 4. The sum of n terms of AP is given by thre formula S_{n} = n/2 [2a + (n - 1) d]

S_{15 }= 15/2 [2 × 4 + (15 - 1)(- 5)]

= 15/2 [8 + 14 (- 5)]

= 15/2 [8 - 70]

= 15/2 × (- 62)

= - 465

**Video Solution:**

## Show that a_{1}, a_{2},... , an , ...form an AP where an is defined as below (i) a_{n} = 3 + 4n (ii) a_{n} = 9 - 5n. Also find the sum of the first 15 terms in each case

### Class 10 Maths NCERT Solutions- Chapter 5 Exercise 5.3 Question 10 :

Show that a1, a2,... , an , ...form an AP where an is defined as below (i) an = 3 + 4n (ii) an = 9 - 5n. Also find the sum of the first 15 terms in each case

If a1, a2,... , an , ...form an AP where an is defined as below then the sum of first 15 terms for general term an = 3 + 4n is equal to 525. and the sum of first terms for general term an = 9 - 5n is equal to -465