Advertisements
Advertisements
प्रश्न
The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.
उत्तर
Let a be the first term and d be the common difference.
We know that, nth term = an = a + (n − 1)d
According to the question,
a5 + a9 = 30
⇒ a + (5 − 1)d + a + (9 − 1)d = 30
⇒ a + 4d + a + 8d = 30
⇒ 2a + 12d = 30
⇒ a + 6d = 15 .... (1)
Also, a25 = 3(a8)
⇒ a + (25 − 1)d = 3[a + (8 − 1)d]
⇒ a + 24d = 3a + 21d
⇒ 3a − a = 24d − 21d
⇒ 2a = 3d
⇒ a = \[\frac{3}{2}\] d ....(2)
Substituting the value of (2) in (1), we get
⇒ 3d + 12d = 15 × 2
⇒ 15d = 30
⇒ d = 2
⇒ a = \[\frac{3}{2}\] [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 5, 7, 9, .... .
APPEARS IN
संबंधित प्रश्न
Find the sum of the following arithmetic progressions:
a + b, a − b, a − 3b, ... to 22 terms
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
If (2p – 1), 7, 3p are in AP, find the value of p.
The first term of an A.P. is p and its common difference is q. Find its 10th term.
How many terms of the series 18 + 15 + 12 + ........ when added together will give 45?
If the second term and the fourth term of an A.P. are 12 and 20 respectively, then find the sum of first 25 terms:
Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20, ......
Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]`
S100 = `square/2 [24 + (100 - 1)"d"]`
= `50(24 + square)`
= `square`
= `square`
Shubhankar invested in a national savings certificate scheme. In the first year he invested ₹ 500, in the second year ₹ 700, in the third year ₹ 900 and so on. Find the total amount that he invested in 12 years
Find the sum of last ten terms of the AP: 8, 10, 12,.., 126.
The sum of 41 terms of an A.P. with middle term 40 is ______.