Advertisements
Advertisements
प्रश्न
The sum of the first n terms in an AP is `( (3"n"^2)/2 +(5"n")/2)`. Find the nth term and the 25th term.
उत्तर
Let S denotes the sum of first n terms of the AP.
∴ `"s"_"n" = ((3"n"^2) /2 +(5"n")/2)`
⇒ ` "s"_("n"-1) = (3("n"-1)^2)/2 + (5 ("n"-1))/2`
`= (3("n"^2 - 2"n" + 1))/2 + (5("n"-1))/2`
`=(3"n"^2 - "n"-2)/2`
∴ nth term pf the AP, an
= `"s"_"n" - "s"_("n"-1)`
= `((3"n"^2 + 5"n")/2) - ((3"n"^2 -"n"-2)/2)`
= `(6"n"+2)/2`
= 3n + 2
Putting n = 25, we get
a25 = 3 × 25 + 1 = 75 + 1 = 76
Hence, the nth term is (3n + 2) and 25th term is 76.
APPEARS IN
संबंधित प्रश्न
If the sum of the first n terms of an A.P. is `1/2`(3n2 +7n), then find its nth term. Hence write its 20th term.
Find the sum given below:
–5 + (–8) + (–11) + ... + (–230)
Find the sum of first 15 multiples of 8.
Find the sum 25 + 28 + 31 + ….. + 100
If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
If the sums of n terms of two arithmetic progressions are in the ratio \[\frac{3n + 5}{5n - 7}\] , then their nth terms are in the ratio
Q.6
Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.