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Arithmetic Progressions

Question
CBSEENMA10009796
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How many terms of an A.P. 9, 17, 25, …. must be taken to give a sum of 636?

Solution

Let there be n terms of this A.P.

For this A.P.,  a = 9

d = a2 - a1 = 17 - 9 = 8

   sn = n22a + (n-1) d⇒ 636 =  n22 x 9 + (n-1) x 8⇒ 636 = n9 + (n-1) x 4⇒ 636 = n ( 9 + 4n - 4 ) ⇒ 636 = n( 4n + 5 )⇒ 636 = 4n2 + 5n⇒ 4n2 + 5n - 636 = 0⇒   4n2 + 53n - 48n - 636 = 0⇒ n ( 4n + 53 ) - 12 ( 4n + 53 ) = 0⇒ ( 4n + 53 )  ( n - 12 ) = 0⇒ 4n + 53 = 0  or  n-12 = 0⇒  n = -534   or  n= 12

Since number of terms can neither be negative nor fractional,

we have  n = 12.

Some More Questions From Arithmetic Progressions Chapter

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a – 10, d = 10

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = –2, d = 0

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
a = 4, d = – 3

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

a = – 1.25, d = – 0.25

For the following APs, write the first term and the common difference
3, 1, – 1, – 3, . . .

For the following APs, write the first term and the common difference
– 5, – 1, 3, 7, . . .

For the following APs, write the first term and the common difference
0.6, 1.7, 2.8, 3.9, . . .

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