1200 1300 1400 ( 100 n 1100 )(iii) In n-numbers of months, he deposits Tk. { 2 ( 1 ) }
2
a n dn
= Tk. { 2 1200 ( 1 ) 100 }
2
u n
n= Tk. ( 2400 100 100 )
2
n
n
= Tk. 2 ( 1150 50 )
2
nn
u = Tk. n( 50 n 1150 )
(iv) We know that 1 year = 12 months. Here n 12.
Therefore, A deposits in 1 year = Tk. 12 ( 50 u 12 1150 )
= Tk. 12 ( 600 1150 )
= Tk. 21 u 0517
= Tk. 21000.
Exercise 13.1
Find the common difference and the 12th terms of the series
2 5 12 19
Which term of the series 8 11 14 17 is 392?
Which term of the series 4 7 10 13 is 301?
If the pth term of an arithmetic series is p^2 and qth term is q^2 , what is
(pq)th term of the series?
If the mth tem of an arithmetic series is n and nth term is m, what is
(mn)th term of the series?
What is the number of n terms of the series 1 3 5 7 ?
What is the sum of first 9 terms of the series 8 16 24 n?
5 11 71 23 95 What?
29 25 21 23 What?
The 12th term of an arithmetic series is 7 7. What is the sum of the first 23 terms?
If the 1 6 th term of an arithmetic series is 20 , what will be the sum of first 31
terms?
The total sum of first n terms of the series 9 7 5 is 144. Find
the value of n.
If the sum of first n terms of the series 2 4 6 8 is 25 50 , find the
value of n.
If the sum of first n terms of the series is n(n 1 ), find the series.
If the sum of first n terms of the series is n(n 1 ), what is the sum of first 1 0
terms?
1 6. If the sum of 12 terms of an arithmetic series is 144 and first 20 terms is 560, find
the sum of first 6 terms.