Question

In class, the teacher asks every student to write an example of an A.P. Two boys, Aryan and Roshan, write the progressions as −5, −2, 1, 4, ... and 187, 184, 181, ... respectively. Now the teacher asks his various students the following questions on progression.

Help the students to find answers for the following:

1. Find the sum of the common difference of two progressions.   (1)
2. Find the 34^th term of the progression written by Roshan.   (1) 
3. 1. Find the sum of the first 10 terms of the progression written by Aryan.  (2)
      OR
   2. Which term of the progression will have the same value?  (2)

Answer 1

First A.P. by Aryan is −5, −2, 1, 4, ....

Here, first term = a = −5

Common difference (d) = (−2) − (−5)

= −2 + 5

= 3

n^th term = a + (n – 1)d

= (–5) + (n – 1) × 3

Second A.P. by Roshan is 187, 184, 181, ....

Here, first term = a = 187

Common difference (d) = 187 − 184

= −3

n^th term = a + (n – 1)d

= 187 + (n – 1) × (–3)

i. Sum of common difference = Common difference of 1^st A.P. + Common difference of 2^nd A.P. 

= 3 + (−3)

= 3 − 3

= 0

ii. We need to find the 34^th term:

Using formula:

a_n = a + (n − 1)d

Putting a = 187, d = −3, n = 34

t₃₄ = 187 + (34 – 1) × (–3)

= 187 + 33 × (–3)

= 187 – 33 × 3

= 187 – 99

t₃₄ = 88

iii. (A) We know that

Sum of A.P. = `n/n [2a + (n - 1) d]`

Putting a = –5, d = 3, n = 10

= `10/2 [2(-5) + (10 - 1)(3)]`

= 5(–10 + 9 × 3)

= 5(–10 + 27)

= 5(17)

= 85

OR

(B) Since both terms of the A.P. have the same value, n^th terms of both would be equal.

(–5) + (n – 1) × 3 = 187 + (n – 1) × (–3)

–5 + 3n – 3 = 187 – 3n + 3

3n + 3n = 187 + 3 + 5 + 3

6n = 198

n = `198/6`

n = 33

Thus the 33^rd term of both A.P. will have the same value.