Advertisements
Advertisements
प्रश्न
In an A.P. (with usual notations) : given d = 5, S9 = 75, find a and a9
Advertisements
उत्तर
d = 5, S9 = 75
an = a + (n – 1)d
a9 = a + (9 – 1) x 5
= a + 40 ...(i)
S9 = `n/(2)[2a + (n - 1)d]`
75 = `(9)/(2)[2a + 8 xx 5]`
`(150)/(9)` = 2a + 40
2a = `(150)/(9) - 40`
= `(50)/(3) - 40`
2a = `(-70)/(3)`
⇒ a = `(-70)/(2 xx 3)`
a = `(-35)/(3)`
From (i),
a9 = a + 40
= `(-35)/(3) + 40`
= `(-35 + 120)/(3)`
= `(85)/(3)`
∴ a = `(-35)/(3), a_9 = (85)/(3)`.
APPEARS IN
संबंधित प्रश्न
The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
In an AP given d = 5, S9 = 75, find a and a9.
If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.
What is the sum of first n terms of the AP a, 3a, 5a, ....
How many terms of the AP 63, 60, 57, 54, ... must be taken so that their sum is 693? Explain the double answer.
Find the first term and common difference for the A.P.
127, 135, 143, 151,...
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d)
For an A.P., if t1 = 1 and tn = 149, then find Sn.
Activitry :- Here t1= 1, tn = 149, Sn = ?
Sn = `n/2 (square + square)`
= `n/2 xx square`
= `square` n, where n = 75
Find the sum of odd natural numbers from 1 to 101.
Find the sum of last ten terms of the AP: 8, 10, 12,.., 126.
