Advertisements
Advertisements
Question
A man saved ₹33000 in 10 months. In each month after the first, he saved ₹100 more than he did in the preceding month. How much did he save in the first month?
Solution
Let the money saved by the man in the first month be ₹a
It is given that in each month after the first, he saved ₹100 more than he did in the preceding month. So, the money saved by the man every month is in AP with common difference ₹100.
∴ d =Rs 100
Number of months, n = 10
Sum of money saved in 10 months, s10 = ₹ 33,000
`"Using the formula ," S_n = n/2 [2a + (n-1) d],`we get
`S_10 = 10/2 [2a + (10-1) xx 100 ] = 33000`
⇒ 5 (2a +900 )= 33000
⇒ 2a +900=6600
⇒ 2a = 6600-900=5700
⇒ a = 2850
Hence, the money saved by the man in the first month is ₹2,850.
APPEARS IN
RELATED QUESTIONS
In an AP. It is given that `s_5 + s_7 = 167 and s_10 = 235 ," then find the AP, where " S_n` denotes the sum of its first n terms.
In an AP, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its common difference.
An AP 5, 12, 19, .... has 50 term. Find its last term. Hence, find the sum of its last 15 terms.
The sum of the 4th and 8th terms of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of its first 10 terms.
The sum of fist m terms of an AP is ( 4m2 - m). If its nth term is 107, find the value of n. Also, Find the 21st term of this AP.
A man arranges to pay off debt of ₹36000 by 40 monthly instalments which form an arithmetic series. When 30 of the installments are paid, he dies leaving on-third of the debt
unpaid. Find the value of the first instalment.
Which of the following sequences are A.P.? If they are A.P. find the common difference.
A = {1, 2, 3,4, 5}, B = {5, 6, 7} write AUB .
Yogesh requires 3 days more than Vivek to do a work completely. If both
of them work together, the work can be completed in 2 days. Find the
number of days required for each of them to do the work completely.
Find tn for the A.P. 3,8,13,18,.....
