Advertisements
Advertisements
प्रश्न
Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the installments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30th installments of loan? What amount of loan she still has to pay after the 30th installment?
Advertisements
उत्तर
Total amount of loan = Rs. 1,18,000
First installment = a = Rs. 1000
Increase in installment every month = d = Rs. 100
30th installment = t30
= a + 29d
= 1000 + 29 × 100
= 1000 + 2900
= Rs. 3900
Now, amount paid in 30 installments = S30
= `30/2 [2 xx 1000 + 29 xx 100]`
= 15[2000 + 2900]
= 15 × 4900
= Rs. 73,500
∴ Amount of loan to be paid after the 30th installments
= Rs. (1,18,000 – 73,500)
= Rs. 44,500
संबंधित प्रश्न
In an AP given d = 5, S9 = 75, find a and a9.
How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?
Find the sum of all 3-digit natural numbers, which are multiples of 11.
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
The 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94. Find the common difference of the AP.
How many two-digits numbers are divisible by 3?
Choose the correct alternative answer for the following question .
First four terms of an A.P. are ....., whose first term is –2 and common difference is –2.
If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is ______.
Q.11
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?
