If the ratio of the sum to n terms of two A ps is then the ratio of their 17th terms is

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The sum of n terms of two arithmetic series are in the ratio2n + 3: 6n + 5, then the ratio of their 13th terms is

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Last updated at Oct. 8, 2021 by Teachoo

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Example 7 The income of a person is Rs. 3,00,000, in the first year and he receives an increase of Rs.10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years. Income of the person in 1st year = Rs 3,00,000 Income of the person in 2nd year = Rs 3,00,000 + 10,000 = Rs 3,10,000 Income of the person in 3rd year = Rs 3,10,000 + 10,000 = Rs 3,20,000 Thus, Income received every year is 300000, 310000, 320000, This is an A.P as difference between consecutive terms is constant. 300000, 310000, 320000, Here first term = a = 300000 common difference = d = 310000 300000 = 10000 To find the total amount received in 20 years, we use the formula Sn = n/2 ( 2a + (n 1)d ) Where, Sn = sum of n terms of A.P. n = number of terms a = first term and d = common difference Here, Sn = Amount received in 20 years, n = 20 , a = 300000, d = 10000 Putting the values S20= 20/2 [2 300000 + (20 1)10000] S20= 20/2 [2 300000 + (20 1)10000] = 10 [ 600000 + 19 10000] = 10 [600000 + 190000] = 10 (790000) = 79,00,000 Thus , the total amount received at the end of 20 years is Rs. 79,00,000.

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression,

`(S_n)/(S'_n) = (3n + 5)/(5n+ 7)`                  ....(1)

We need to find the ratio of their nth terms.

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

So,

`S_n = n/2 [ 2a + ( n - 1)d]`

Where, a and d are the first term and the common difference of the first A.P.

Similarly,

`S'_n = n/2 [ 2a' + ( n - 1)d']`

Where, a’ and d’ are the first term and the common difference of the first A.P.

So,

`(S_n)/(S'_n) = (n/2[2a + (n-1)d])/(n/2[2a'+(n-1)d'])`

       `= ([2a + (n-1)d])/([2a'+(n-1)d'])`              ............(2)

Equating (1) and (2), we get,

`= ( [2a + (n-1)d])/( [2a'+(n-1)d'])=(3n + 5)/(5n + 7)`

Now, to find the ratio of the nth term, we replace n by2n - 1 . We get,

`= ( [2a + (n-1-1)d])/( [2a'+(n-1-1 )d'])= (3(2n - 1) + 5)/(5(2n -1) + 7)`

` ( 2a + (2n-2)d)/( 2a'+(2n-2)d')= (6n - 3 + 5 )/(10n - 5 + 7)`

` ( 2a + 2(n-1)d)/( 2a'+2(n-1)d')=(6n +2)/(10n + 2)`

` ( a + (n-1)d)/( a'+(n-1)d') = (3n + 1)/(5n + 1)`

As we know,

an = a + (n - 1)d 

Therefore, we get,

`a_n /(a'_n) = (3n + 1) /(5n +1)`