Question

Sum of the first 4 terms of an arithmatic sequence is 72.Sum of the first 9 terms of the arithmatic sequence is also 72 ​

Answer 1

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☆ Given -

Sum of first 4 terms = 72

Sum of first 9 terms = 72

☆ To Find -

The sequence..

☆ Solution -

Let the first term = a

Common Difference = d

Now ,

Using the Formula for Sum of first n terms

$\rm \: S_{n} \: = \frac{n}{2} (2a + (n - 1)d)$

Now , put n = 4

$\longrightarrow \rm \: S_{4} = \frac{4}{2} (2a + (4 - 1)d)$

$\longrightarrow \: \rm72 = 2(2a + 3d) \\ \\ \implies \rm\: 2a + 3d = 36....(i)$

Now , put n = 9

$\longrightarrow \rm \: S_{9} = \frac{9}{2} (2a + (9 - 1)d)$

$\implies \rm \: 72 = \frac{9}{2}(2a + 8d) \\ \\ \implies \: \rm \: a + 4d = 8 \\ \\ \implies \: \rm \: a = 8 - 4d ....(ii)$

Put the value of a from eq(ii) in eq(i)

$\implies \: 2(8 - 4d) + 3d = 36$

$\implies \: 16 - 8d + 3d = 36$

$\implies \: - 5d = 36 - 16$

$\implies \: - 5d = 20$

$\implies \: d = - 4$

put the value of d in equation (ii)

$\implies \: a = 8 - 4d$

$\implies \: a = 8 - 4( - 4) \\ \\ \implies \: a = 8 + 16 = 24$

So , sequence would be -

a , a+d , a+2d , a+3d,.....

24 , 20 , 16 , 12 , 8 , 4 ,.....