Question

Find the A.P. whose 1st term is 100 and the sum of the first six terms is five times the sum of next six terms?

Answer 1

a1+d=a2

a2 + d =a3

100 +(-10) = 90

90 + (-10)= 80

80+(-10) =70

AP : 100,90,80,70

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Answer 2

Answer :-

The A.P. is - 100, 90, 80, 70, 60.

Information we get from the question -

First term (a) = 100

Sum of the first six terms ($S_6$)

= 5 * [Sum of next six terms]

= 5 * [Sum of 12 terms - Sum of first 6 terms]

Solution :-

Sum of n terms ($S_n$) = $\dfrac{n}{2} (2a+(n-1)d)$

where,

d = the difference between two consecutive terms of the A.P.

By using this formula,  we will find sum of the first six terms ($S_6$)

$\Longrightarrow S_6=\dfrac{6}{2} (2\times 100+(6-1)d)\\\\\Longrightarrow S_6=\dfrac{6}{2} (2\times 100+(5)d)\\\\\Longrightarrow S_6=\dfrac{6}{2} (200+5d)\\\\\Longrightarrow S_6=3(200+5d)\\\\\Longrightarrow S_6=600+15d$

Now, we will find sum of the first 12 terms ($S_{12}$)

$\Longrightarrow S_{12}=\dfrac{12}{2} (2\times 100+(12-1)d)\\\\\Longrightarrow S_{12}=\dfrac{12}{2} (2\times 100+(11)d)\\\\\Longrightarrow S_{12}=\dfrac{12}{2} (200+11d)\\\\\Longrightarrow S_{12}=6 (200+11d)\\\\\Longrightarrow S_{12}=1200+66d$

According to the Question -

Sum of the first six terms ($S_6$) = 5 * [Sum of 12 terms - Sum of first 6 terms]

$S_6$ = 5 * [$S_{12}$ - $S_6$]

$\Longrightarrow 600+15d =5\times [(1200+66d)-(600+15d)]\\\\\Longrightarrow 600+15d =5\times [1200+66d-600-15d]\\\\\Longrightarrow 600+15d =5\times [600+51d]\\\\\Longrightarrow 600+15d =3000+255d\\\\\Longrightarrow -255d+15d =3000-600\\\\\Longrightarrow -240d =2400\\\\\Longrightarrow d =\dfrac{2400}{-240}\\\\\Longrightarrow d =-10$

Finding the A.P. till 5 terms.

1st term = a = 100

2nd term = a + (n-1)d = 100 + (2-1)-10 = 100 + (1)-10 = 100 - 10 = 90

3rd term = a + (n-1)d = 100 + (3-1)-10 = 100 + (2)-10 = 100 - 20 = 80

4th term = a + (n-1)d = 100 + (4-1)-10 = 100 + (3)-10 = 100 - 30 = 70

5th term = a + (n-1)d = 100 + (5-1)-10 = 100 + (4)-10 = 100 - 40 = 60

The A.P. is - 100, 90, 80, 70, 60.