Advertisements
Advertisements
प्रश्न
In an A.P., if the first term is 22, the common difference is −4 and the sum to n terms is 64, find n.
उत्तर
In the given problem, we need to find the number of terms of an A.P. Let us take the number of terms as n.
Here, we are given that,
a = 22
d = -4
S_n= 6
So, as we know the formula for the sum of n terms of an A.P. is given by,
`S_n = n/2 [2a + (n - 1)d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula we get,
`S_n= n/2 [2(22) + (n - 1)(-4)]`
`64 = n/2[44 - 4n + 4]`
64(2) = n(48 - 4n)
`128 = 48n - 4n^2`
Further rearranging the terms, we get a quadratic equation,
`4n^2 - 48n + 128 = 0`
On taking 4 common we get
`n^2 - 12n + 32 = 0`
Further, on solving the equation for n by splitting the middle term, we get,
`n^2 - 12n + 32 = 0`
`n^2 - 8n -4n + 32 = 0`
n(n - 8) - 4(n - 8) = 0
(n - 8)(n - 4) = 0
So, we get,
(n - 8) = 0
n = 8
Also
(n - 4) = 0
n = 4
Therefore n = 4 or 8
APPEARS IN
संबंधित प्रश्न
The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`
A sum of ₹2800 is to be used to award four prizes. If each prize after the first is ₹200 less than the preceding prize, find the value of each of the prizes
If the common differences of an A.P. is 3, then a20 − a15 is
If the sum of first n terms of an A.P. is \[\frac{1}{2}\] (3n2 + 7n), then find its nth term. Hence write its 20th term.
If the sum of first p term of an A.P. is ap2 + bp, find its common difference.
Sum of n terms of the series `sqrt2+sqrt8+sqrt18+sqrt32+....` is ______.
If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
The 11th term and the 21st term of an A.P are 16 and 29 respectively, then find the first term, common difference and the 34th term.
Complete the following activity to find the 19th term of an A.P. 7, 13, 19, 25, ........ :
Activity:
Given A.P. : 7, 13, 19, 25, ..........
Here first term a = 7; t19 = ?
tn + a + `(square)`d .........(formula)
∴ t19 = 7 + (19 – 1) `square`
∴ t19 = 7 + `square`
∴ t19 = `square`
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.