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Question
How many terms are present in the sequence of A.P. 6, 11, 16, 21, ......... whose sum is 969?
Solution
Given sequence is 6, 11, 16, 21, .....
Let a be the first term and d be the common difference.
Then, a = 6, d = 11 – 6 = 5
We know, `S_n = n/2 [2a + (n - 1)d]`
⇒ 969 = `n/2 [2(6) + (n - 1) (5)]`
⇒ 969 = `n/2 [12 + 5n - 5]`
⇒ 969 = `n/2 [7 + 5n]`
⇒ 1938 = 7n + 5n2
⇒ 5n2 + 7n – 1938 = 0
⇒ 5n2 + (102 – 95)n – 1938 = 0
⇒ 5n2 + 102n – 95n – 1938 = 0
⇒ n(5n + 102) – 19(5n + 102) = 0
⇒ (5n + 102) (n – 19) = 0
n = 19, – `102/5`
The negative value is rejected because the number of words cannot be negative.
As a result, the A.P. contains 19 terms.
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