Advertisements
Advertisements
प्रश्न
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
Advertisements
उत्तर
Given AP is
3, 15, 27, 39 .....
where a = 3, d = 15 - 3 = 12
Let the nth term be 120 more than its 21st term.
tn = t21 + 120
⇒ 3 +(n -1 )12 = 3 +20 × 12 + 120
⇒ (n - 1) × 12 = 363 - 3
⇒ (n - 1) =`360/12`
∴ n = 31
Hence , the required term is
t31 = 3 + 30 × 12
= 363
APPEARS IN
संबंधित प्रश्न
Ramkali required Rs 2,500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.
What value is generated in the above situation?
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
In an A.P., the sum of first n terms is `(3n^2)/2 + 13/2 n`. Find its 25th term.
The sum of n natural numbers is 5n2 + 4n. Find its 8th term.
The sum of the first n terms of an AP is (3n2 + 6n). Find the nth term and the 15th term of this AP.
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
First four terms of the sequence an = 2n + 3 are ______.
The famous mathematician associated with finding the sum of the first 100 natural numbers is ______.
