the sum of the first 7 terms of an arithmetic progression is 140 and the sum of the next 7 term of the sum progression is 385 then find the arithmetic progression
Answers
Answer:
Solution :-
Sum of first 7 terms (S_{7}S
7
) = 140
Sum of next 7 terms = 385
So, Sum of 14(7+7) terms (S_{14}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]S
n
=
2
n
[2a+(n−1)d]
where,
S_{n}S
n
= 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}
=>S
7
=
2
7
[2a+(7−1)d]
=>140=
2
7
[2a+6d]
=>
7
140×2
=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}
=>S
14
=
2
14
[2a+(14−1)d]
=>525=7[2a+13d]
=>
7
525
=2a+13d
=>75=2a+13d
Subsititutingthevalueofafromeq.i.)
=>75=40−6d+13d
=>75−40=13d−6d
=>35=7d
=>d=
7
35
=>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}
2
10
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.