Advertisements
Advertisements
Question
An article can be bought by paying Rs. 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs. 3,000 and every other installment is Rs. 100 less than the previous one, find:
- amount of installments paid in the 9th month.
- total amount paid in the installment scheme.
Advertisements
Solution
Number of installments = n = 12
First installment = a = Rs. 3000
Depreciation in installment = d = –100
i Amount of installment paid in the 9th month
= t9
= a + 8d
= 3000 + 8 × (–100)
= 3000 – 800
= Rs. 2200
ii. Total amount paid in the installment scheme
= S12
= `12/2 [2 xx 3000 + 11 xx (-100)]`
= 6[6000 – 1100]
= 6 × 4900
= Rs. 29,400
RELATED QUESTIONS
Find the sum of the following APs:
2, 7, 12, ..., to 10 terms.
Find the sum of first 15 multiples of 8.
In a flower bed, there are 43 rose plants in the first row, 41 in second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?
Find the sum of the following Aps:
i) 2, 7, 12, 17, ……. to 19 terms .
The first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum?
Q.1
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.
The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 equal to the sum of first 2n terms of another A.P. whose first term is – 30 and the common difference is 8. Find n.
Find the sum:
`(a - b)/(a + b) + (3a - 2b)/(a + b) + (5a - 3b)/(a + b) +` ... to 11 terms
Kanika was given her pocket money on Jan 1st, 2008. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?
