मराठी

How Many Terms of the A.P. 63, 60, 57, ... Must Be Taken So that Their Sum is 693? - Mathematics

Advertisements
Advertisements

प्रश्न

How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?

Advertisements

उत्तर

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.

A.P. 63, 60, 57

So here, let us find the number of terms whose sum is 693. For that, we will use the formula,

`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.

= number of terms

The first term (a) = 63

The sum of n terms (Sn) = 693

Common difference of the A.P. (d) = `a_2 - a_1`

= 60 - 63

= -3

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

`639 = n/2 [2(63) + (n -2)(-3)]`

`693 = (n/2) [126 + (-3n + 3)]`

`693 = (n/2) [129 - 3n`]

`693(2) = 129n - 3n^2`

So, we get the following quadratic equation,

`3n^2 - 129n + 1386 = 0`

`n^2 - 43n + 462 = 0`

On solving by splitting the middle term, we get,

`n^2 - 22n - 21n + 462 = 0`

`n(n - 22) - 21(n - 22) = 0`

(n - 22)(n - 21) = 0

Further,

n - 22 =0 

n = 22

or

n - 21 = 0

n = 21

Here 22 nd term wil be

`a_22 = a_1 + 21d`

= 63 + 21(-3)

= 63 - 63

= 0

So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (n) is 21 or 22

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Arithmetic Progression - Exercise 5.6 [पृष्ठ ५१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 10
पाठ 5 Arithmetic Progression
Exercise 5.6 | Q 10.4 | पृष्ठ ५१

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.


How many terms of the series 54, 51, 48, …. be taken so that their sum is 513 ? Explain the double answer


The first and last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?


Find the sum of the following arithmetic progressions:

a + b, a − b, a − 3b, ... to 22 terms


Find the sum of all integers between 84 and 719, which are multiples of 5.


The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.


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


The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.


For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... equal?


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


In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to


The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is 

 

Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the installments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30th installments of loan? What amount of loan she still has to pay after the 30th installment?


Q.1


Q.5 


If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.


In an A.P. (with usual notations) : given d = 5, S9 = 75, find a and a


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)


If the first term of an A.P. is 5, the last term is 15 and the sum of first n terms is 30, then find the value of n.


The nth term of an A.P. is 6n + 4. The sum of its first 2 terms is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×