English

If 5 + 9 + 13 + . . . to N Terms 7 + 9 + 11 + . . . to (N + 1) Terms = 17 16 , Then N = - Mathematics

Advertisements
Advertisements

Question

If \[\frac{5 + 9 + 13 + . . . \text{ to n terms} }{7 + 9 + 11 + . . . \text{ to (n + 1) terms}} = \frac{17}{16},\] then n =

 

Options

  • 8

  • 7

  • 10

  • 11

MCQ
Advertisements

Solution

Here, we are given,

\[\frac{5 + 9 + 13 + . . . \text{ to n terms} }{7 + 9 + 11 + . . . \text{ to (n + 1) terms}} = \frac{17}{16},\]             ...........(1) 

We need to find n.

So, first let us find out the sum of n terms of the A.P. given in the numerator ( 5 + 9 + 13 + ...) . Here we use the following formula for the sum of n terms of an A.P.,

`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

Here,

Common difference of the A.P. (d) =  a2 - a

= 9 - 5

= 4

 Number of terms (n) = n

First term for the given A.P. (a) = 5 

So, using the formula we get,

`S_n = n/2 [ 2(5) + ( n-1) ( 4) ] `

     `= (n/2) [ 10 + (4n - 4)] `

     ` = ( n/2) (6 +4n) ` 

      = n (3 + 2n)                ....(2) 

Similarly, we find out the sum of ( n + 1)  terms of the A.P. given in the denominator ( 7 + 9 + 11+...) .

Here,

Common difference of the A.P. (d) = a2- a1 

= -9 - 7

= 2 

Number of terms (n) = n

First term for the given A.P. (a) = 7 

So, using the formula we get,

`S_(N+1) = ( n+1)/2 [ 2(7) + [( n + 1 ) - 1](2)]`

        `= ((n + 1) /2 )[14 + ( n) ( 2 ) ] `

        = ( n + 1 ) ( 7 + n) 

        = 7n + 7 + n2 + n 

        = n2 + 8n + 7              ..........(3) 

Now substituting the values of (2) and (3) in equation (1), we get,

     `(2n^2 + 3n ) /(n^2 + 8n + 7) = 17/16`

                         16 (2n2 + 3n ) = 17 ( n2+ 8n + 7)

                              32n2 + 48n = 17n2+ 136n + 119

32n2 - 17n2 + 48n - 136n - 119 = 0

                       15n2 - 88n - 119 = 0

 

Further solving the quadratic equation for n by splitting the middle term, we get,

             15n2 - 88n - 119 = 0

15n2 - 105n + 17n +- 119 = 0

    15n ( n - 7) + 17 ( n - 7) = 0

               ( 15n + 17 )(n- 7) = 0

So, we get

15n + 17 = 0

       15n = - 17

          `n = (-17)/15`

Or

n - 7 = 0

     n = 7 

Since is a whole number, it cannot be a fraction. So, n = 7 

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Arithmetic Progression - Exercise 5.8 [Page 59]

APPEARS IN

RD Sharma Mathematics [English] Class 10
Chapter 5 Arithmetic Progression
Exercise 5.8 | Q 32 | Page 59

RELATED QUESTIONS

If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.


The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1)


Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.


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.


The 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94. Find the common difference of the AP. 


Determine k so that (3k -2), (4k – 6) and (k +2) are three consecutive terms of an AP.


Divide 24 in three parts such that they are in AP and their product is 440.


Find the sum of the following Aps:

i) 2, 7, 12, 17, ……. to 19 terms . 


If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.

 

If \[\frac{1}{x + 2}, \frac{1}{x + 3}, \frac{1}{x + 5}\]  are in A.P. Then, x =


If 18, ab, −3 are in A.P., the a + b =


In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms. 


Find the sum of first 10 terms of the A.P.

4 + 6 + 8 + .............


How many terms of the series 18 + 15 + 12 + ........ when added together will give 45?


In an A.P. sum of three consecutive terms is 27 and their products is 504. Find the terms. (Assume that three consecutive terms in an A.P. are a – d, a, a + d.)


Find the sum of numbers between 1 to 140, divisible by 4


The famous mathematician associated with finding the sum of the first 100 natural numbers is ______.


If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are ______.


Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..


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×