Q) The ratio of the 11th term to 17th term of an A.P. is 3:4. Find the ratio of 5th term to 21st term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.

Ans:

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STEP BY STEP SOLUTION

We know that nth term of an A.P.  =  a + (n-1) d

Therefore,  11th term, N11 = a + 10d and 17th term, N17 = a + 16d

Given that : \frac{N_1_1}{N_1_7} = \frac{3}{4}

\therefore    \frac{a+10d}{a+16d} = \frac{3}{4}

or 4a + 40 d = 3a + 48 d

or a = 8d ……………………………………. equation no. (i)

Now 5th term N5 = a + 4d and 21st term N21 = a + 20 d

Therefore,  \frac{N_5}{N_2_1} = \frac{a+4d}{a+20d}

Substituting a = 8d from equation (i), we get

\frac{8d+4d}{8d+20d}   = \frac{12d}{28d} = \frac{3}{7}

Therefore, N5:N21 = 3:7

Let’s calculate ratio of the sum of first 5 terms to that of first 21 terms.

We know that sum of n terms of an A.P.  Sn = \frac{n}{2} (2a + (n-1) d)

Sum of first 5 terms, S5 = \frac{5}{2}(2a + 4d)

and Sum of first 21 terms, S21 =    \frac{21}{2} (2a + 20d)

Therefore, \frac{S_5}{S_2_1} = \frac{\frac{5}{2}(2a+4d)}{\frac{21}{2}(2a+20d)}

= \frac{5(2a+4d)}{21(2a+20d)}

Substituting a = 8d from equation (i), we get

= \frac{5(16d+4d)}{21(16d+20d)}

= \frac{5(20d)}{21(36d)}

= \frac{25}{189}

Therefore, S5 : S21 = 25 : 189

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