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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​

Answers

Answered by mridulmilansarkar
3

Solution

Sum of first 7 terms (S_{7}S7 ) = 140

Sum of next 7 terms = 385

So, Sum of 14(7+7) terms (S_{14}S14 )= 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]Sn=2n[2a+(n−1)d]

where,

S_{n}Sn = Sum of n terms

n = Number of terms

a = First term

d = difference of consecutive terms

So,

\begin{gathered}= > 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.)\end{gathered}=>S7=27[2a+(7−1)d]=>140=27[2a+6d]=>7140×2=2a+6d=>40=2a+6d=>40−6d=2a....i.)

Now,

\begin{gathered}= > 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\end{gathered}=>S14=214[2a+(14−1)d]=>525=7[2a+13d]=>7525=2a+13d=>75=2a+13dSubsititutingthevalueofafromeq.i.)=>75=40−6d+13d=>75−40=13d−6d=>35=7d=>d=735=>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}210

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.

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