# Question 1) Find the sum of the first 24 terms of the list of numbers whose nth term is given (3+2n). Question 2) For what value of n , are the nth terms of the two AP 63,65,67 and 3,10,17........ are equal ? Question 3) The pth term of an AP is a and qth term is b . Prove that the sum of its (p+q) terms = p+q/2{ a+ b+ (a-b/p-q)}.

### Asked by arindeep.singh | 22nd Jul, 2020, 06:24: PM

(1)

a_{n}= 3+2n

Now, put n=1,2,3,...

a_{1}= 3+2(1) = 5

a_{2}= 3+2(2) = 7

a_{3}= 3+2(3) = 9

Thus, the terms of the AP are 5,7,9,....

Here, a = 5 and d = 2

S_{24}= [25+(24 -1)2]

= 12[10+46]

= 12 56

= 672

(2)

In the first AP, a = 63 and d = 2

In the second AP, a = 3 and d = 7

As the nth term of both APs be equal

63+(n-1)2 = 3+(n-1)7

63+2n-2 = 3+7n-7

61+2n = 7n-4

5n = 65

n = 13

Hence, the 13th term of the APs are equal.

3)

pth term is a and qth term is b

A + (p - 1)D = a ... (1)

A + (q - 1)D = b ... (2)

Subtracting (2) from (1), we get

(p-1-q+1)D = a-b

D = a-b/p-q

Adding (1) and (2), we get

2A + D(p+q-2) = a+b

2A + D(p+q-1) = a+b+D

### Answered by Renu Varma | 23rd Jul, 2020, 12:32: PM

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