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
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.
n = number of terms
Here,
Common difference of the A.P. (d) = a2 - a1
= 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 n is a whole number, it cannot be a fraction. So, n = 7
APPEARS IN
RELATED QUESTIONS
In an AP, given a = 7, a13 = 35, find d and S13.
In an AP given a = 8, an = 62, Sn = 210, find n and d.
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.
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
If k,(2k - 1) and (2k - 1) are the three successive terms of an AP, find the value of k.
Choose the correct alternative answer for the following question .
In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
Find the sum of n terms of the series \[\left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + . . . . . . . . . .\]
There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.
If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 − S10)
If `4/5` , a, 2 are three consecutive terms of an A.P., then find the value of a.
If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to
x is nth term of the given A.P. an = x find x .
Q.3
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
If the second term and the fourth term of an A.P. are 12 and 20 respectively, then find the sum of first 25 terms:
In an A.P. (with usual notations) : given a = 8, an = 62, Sn = 210, find n and d
Shubhankar invested in a national savings certificate scheme. In the first year he invested ₹ 500, in the second year ₹ 700, in the third year ₹ 900 and so on. Find the total amount that he invested in 12 years
Solve the equation:
– 4 + (–1) + 2 + 5 + ... + x = 437
