Which term of the A.P. 3, 14, 25, 36, … will be 77 more than its 30th term? Find the term.
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Step-by-step explanation:
In the given AP, the first term = a = 3
Common difference = d = 14 – 3 = 11
To find: place of the term which is 99 more than its 25th term.
So, we first find its 25th term.
Since, we know that
an = a + (n-1) × d
∴ a25 = 3 + (25 -1) × 11
⇒ a25 = 3 + 24 × 11
⇒ a25 = 3 + 264
⇒ a25 = 267
∴ 25th term of the AP is 267.
Now, 99 more than 25th term of the AP is 99 + 267 = 366.
So, to find: place of the term 366.
So, let an = 366
Since, we know that
an = a + (n-1) × d
∴ 366 = 3 + (n -1) × 11
⇒ 366 – 3 = 11n - 11
⇒ 363 = 11n - 11
⇒ 363 + 11 = 11n
⇒ 11n = 374
⇒ n = 374/11 = 34
∴ 34th term of the AP is the term which is 99 more than 25th term.
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