Question

The sum of the first 7 terms of an arithmetic progression is 140 and the sum of the

next 7 terms of the same progression is 385 then find the arithmetic progression.​

Answer 1

Solution :-

Sum of first 7 terms ($S_{7}$) = 140

Sum of next 7 terms = 385

So, Sum of 14(7+7) terms ($S_{14}$)= 140 + 385 = 525

We know the formula of the sum of n terms of an A.P. i.e.,

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

where,

$S_{n}$ = Sum of n terms

n = Number of terms

a = First term

d = difference of consecutive terms

So,

$=>S_{7} =\frac{7}{2} [2a+(7-1)d]\\\\=>140 =\frac{7}{2} [2a+6d]\\\\=>\frac{140 \times 2}{7} = 2a+6d\\\\=>40 = 2a + 6d\\\\=>40-6d = 2a\:\:\:\:....i.)$

Now,

$=>S_{14} =\frac{14}{2} [2a+(14-1)d]\\\\=>525 =7 [2a+13d]\\\\=>\frac{525}{7} = 2a+13d\\\\=>75 = 2a + 13d\\\\Subsitituting\:\:the\:\:value\:\:of\:\:a\:\:from\:\:eq.\:\:i.)\\\\=>75 =40-6d+ 13d\\\\=>75-40=13d-6d\\\\=>35=7d\\\\=>d=\frac{35}{7}\\\\=>d=5$

We have the value of d and by subsituting the value of d in eq. i.), we will find the value of a.

=> 2a = 40 - 6d

=> 2a = 40 - 6 x 5

=> 2a = 40 - 30

=> 2a = 10

=> a = $\frac{10}{2}$

The first term of A.P. = a = 5

The second term of A.P. = a + d= 5 + 5 = 10

The third term of A.P. = a + 2d= 5 + 2 x 5 = 5 + 10 = 15

The fourth term of A.P. = a + 3d = 5 + 3 x 5 = 5 + 15 = 20

The fifth term of A.P. = a + 4d = 5 + 4 x 5 = 5 + 20 = 25

The sixth term of A.P. = a + 5d= 5 + 5 x 5 = 5 + 25 = 30

The seventh term of A.P. = a + 6d = 5 + 6 x 5 = 5 + 30 = 35

The eighth term of A.P. = a + 7d = 5 + 7 x 5 = 5 + 35 = 40

The ninth term of A.P. = a + 8d = 5 + 8 x 5 = 5 + 40 = 45

The tenth term of A.P. = a + 9d = 5 + 9 x 5 = 5 + 45 = 50

The eleventh term of A.P. = a + 10d = 5 + 10 x 5 = 5 + 50 = 550

The twelfth term of A.P. = a + 11d = 5 + 11 x 5 = 5 + 55 = 60

The thirteen term of A.P. = a + 12d = 5 + 12 x 5 = 5 + 60 = 65

The fourteen term of A.P. = a + 13d = 5 + 13 x 5 = 5 + 65 = 75

∴ The A.P. -

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75.

Answer 2

Answer:

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