Question

The sum of first 6 term of an AP is 63. The ratio of its 10th term to its 20th term is
1:2. Calculate the first and 15th terms.
​

Answer 1

Answer:

a=3

15TH TERM... 45

Step-by-step explanation:

as we know that,

a+9d/a+19d =1/2 (since, given in question)

2a+18d=a+19d

a=d.....(1)

Now,

n/2[2a+(n-1)d]=63

3[2a+5a]=63

7a=21

a=3

Now, 15th term...

an=a+(n-1)d

=3+14×3

=45

Answer 2

Given:-

Sum of 6 terms of A.P = 63

10th term :20th term = 1:2

To Find :-

First term of the A.P

15th term of the A.P

Formula Used :-

$Tn = a + (n - 1)d$

$Sn = \frac{n}{2} (2a + (n - 1)d)$

Solution :-

$\frac{ \:T10} {T20} = \frac{1}{2} \\ \frac{a + 9d}{a + 19d} = \frac{1}{2} \\ 2a + 18d = a + 19d$

a = d__________Equation 1

Now, by using equation 1 we have

$S6 = \frac{6}{2} (2a + (6 - 1)d) \\ 63 = 3(2a + 5d) \\ 63 = 3(7a) \\ 63 = 21a \\ 3 = a$

And so, d = 3.

$Hence, \: the \: 15th \: term =a+(15-1)d \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 3 + 14 \times 3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 45$