Advertisements
Advertisements
प्रश्न
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
विकल्प
24th term
27th term
26th term
25th term
Advertisements
उत्तर
Here, the sum of first n terms is given by the expression,
`S_n = 3n^2 + 5n`
We ned to find which term of the A.P. is 164.
Let us take 164 as the nth term.
So we know that the nthterm of an A.P. is given by,
`a_n = S_n - S_( n-1) `
So,
`164 = S_n - S_( n-1) `
`164 = 3n^2 + 5n - [ 3(n-1)^2 + 5(n - 1 ) ]`
Using the property,
`( a - b)^2 = a^2 + n^2 - 2ab`
We get,
`164 = 3n^2 + 5n - [3 (n^2 + 1 - 2n ) + 5 (n -1)]`
`164 = 3n^2 + 5n - [ 3n^2 + 3 - 6n + 5n - 5 ]`
`164 = 3n^2 + 5n - (3n^2 - n - 2)`
`164 = 3n^2 + 5n - 3n^2 + n + 2 `
164 = 6n + 2
Further solving for n, we get
6n = 164 - 2
` n = 162/6`
n = 27
Therefore, 164 is the 27th term of the given A.P.
APPEARS IN
संबंधित प्रश्न
If Sn1 denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
Find the sum of the following APs.
−37, −33, −29, …, to 12 terms.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 9 − 5n
Also, find the sum of the first 15 terms.
Write an A.P. whose first term is a and common difference is d in the following.
a = –19, d = –4
If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term?
The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.
If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero.
Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....
If the sum of n terms of an A.P. is 2n2 + 5n, then its nth term is
If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is
If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
If 18, a, b, −3 are in A.P., the a + b =
Q.6
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4.
Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
Measures of angles of a triangle are in A.P. The measure of smallest angle is five times of common difference. Find the measures of all angles of a triangle. (Assume the measures of angles as a, a + d, a + 2d)
The sum of n terms of an A.P. is 3n2. The second term of this A.P. is ______.
