हिंदी

The Sum of N Terms of an A.P. is 3n2 + 5n, Then 164 is Its - Mathematics

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

MCQ
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. 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Arithmetic Progression - Exercise 5.8 [पृष्ठ ५९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 10
अध्याय 5 Arithmetic Progression
Exercise 5.8 | Q 33 | पृष्ठ ५९

संबंधित प्रश्न

In an A.P., if S5 + S7 = 167 and S10=235, then find the A.P., where Sn denotes the sum of its first n terms.


The sum of n, 2n, 3n terms of an A.P. are S1 , S2 , S3 respectively. Prove that S3 = 3(S2 – S1 )


The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.


A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.


Find the sum of all 3-digit natural numbers, which are multiples of 11.


Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 − 3n.


Find the sum of the first 22 terms of the A.P. : 8, 3, –2, ………


Find the 6th  term form the end of the AP 17, 14, 11, ……, (-40).


How many three-digit natural numbers are divisible by 9?


Find the sum of  the following Aps:

9, 7, 5, 3 … to 14 terms


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.


If Sn denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).

 

If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 − S10)

 

Write the expression of the common difference of an A.P. whose first term is a and nth term is b.


Q.13


If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.


Find the sum:

`4 - 1/n + 4 - 2/n + 4 - 3/n + ...` upto n terms


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`


Find the sum of all even numbers from 1 to 250.


Sum of 1 to n natural number is 45, then find the value of n.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×