Advertisements
Advertisements
प्रश्न
A merchant borrows ₹ 1000 and agrees to repay its interest ₹ 140 with principal in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 10. Find the amount of the first instalment
उत्तर
The installments are in A.P.
Amount repaid in 12 instalments (S12)
= Amount borrowed + total interest
= 1000 + 140
∴ S12 = 1140
Number of instalments (n) = 12
Each instalment is less than the preceding one by ₹ 10.
∴ d = – 10
Now, Sn = `"n"/2 [2"a" + ("n" - 1)"d"]`
∴ S12 = `12/2 [2"a" + (12 - 1)(- 10)]`
∴ 1140 = 6[2a + 11(– 10)]
∴ 1140 = 6(2a – 110)
∴ `1140/6` = 2a – 110
∴ 190 = 2a – 110
∴ 2a = 300
∴ a = `300/2` = 150
∴ The amount of first instalment is ₹ 150.
संबंधित प्रश्न
Find the sum of the following APs:
2, 7, 12, ..., to 10 terms.
Find the sum of the first 51 terms of the A.P: whose second term is 2 and the fourth term is 8.
Find the sum 3 + 11 + 19 + ... + 803
How many numbers are there between 101 and 999, which are divisible by both 2 and 5?
In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
What is the sum of first 10 terms of the A. P. 15,10,5,........?
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.
A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by \[\frac{l^2 - a^2}{k - (l + a)}\] , then k =
If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
Q.1
Q.18
In a Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find that:
(i) first term
(ii) common difference
(iii) sum of the first 20 terms.
If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
In an A.P. sum of three consecutive terms is 27 and their products is 504. Find the terms. (Assume that three consecutive terms in an A.P. are a – d, a, a + d.)
Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.
[Hint (iii) : These numbers will be : multiples of 2 + multiples of 5 – multiples of 2 as well as of 5]
Find the sum of first 'n' even natural numbers.
If the first term of an A.P. is p, second term is q and last term is r, then show that sum of all terms is `(q + r - 2p) xx ((p + r))/(2(q - p))`.