Advertisements
Advertisements
प्रश्न
A man starts repaying a loan as first instalment of Rs 100 = 00. If he increases the instalments by Rs 5 every month, what amount he will pay in the 30th instalment?
Advertisements
उत्तर
Let
\[a_{30}\] be the amount a man repays in the 30th instalment.
Let d be the common increment in his instalment every month.
Let a be the initial repayment.
Here, a = 100, d = 5, n = 30
Amount to be repaid in the 30th instalment:
\[a_{30}\]
\[\Rightarrow\] a+ \[\left( n - 1 \right)\] d
\[= 100 + 29 \times 5\]
\[ = 245\]
Hence, the man repays Rs 245 in his 30th instalment.
संबंधित प्रश्न
Find the sum to n terms of the A.P., whose kth term is 5k + 1.
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nthterm is (2m – 1): (2n – 1)
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5:9. Find the value of m.
Show that the following sequence is an A.P. Also find the common difference and write 3 more terms in case.
−1, 1/4, 3/2, 11/4, ...
Find:
18th term of the A.P.
\[\sqrt{2}, 3\sqrt{2}, 5\sqrt{2},\]
Find:
nth term of the A.P. 13, 8, 3, −2, ...
Is 68 a term of the A.P. 7, 10, 13, ...?
The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
Find the 12th term from the following arithmetic progression:
1, 4, 7, 10, ..., 88
The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.
If < an > is an A.P. such that \[\frac{a_4}{a_7} = \frac{2}{3}, \text { find }\frac{a_6}{a_8}\].
Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.
Find the sum of the following arithmetic progression :
3, 9/2, 6, 15/2, ... to 25 terms
Find the sum of the following arithmetic progression :
41, 36, 31, ... to 12 terms
Find the sum of the following arithmetic progression :
a + b, a − b, a − 3b, ... to 22 terms
Find the sum of the following serie:
2 + 5 + 8 + ... + 182
Find the sum of first n natural numbers.
Find the sum of all odd numbers between 100 and 200.
Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.
The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17. Find the progression.
If the sum of a certain number of terms of the AP 25, 22, 19, ... is 116. Find the last term.
If S1 be the sum of (2n + 1) terms of an A.P. and S2 be the sum of its odd terms, then prove that: S1 : S2 = (2n + 1) : (n + 1).
A man saves Rs 32 during the first year. Rs 36 in the second year and in this way he increases his savings by Rs 4 every year. Find in what time his saving will be Rs 200.
A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.
If m th term of an A.P. is n and nth term is m, then write its pth term.
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
If the sum of n terms of an A.P., is 3 n2 + 5 n then which of its terms is 164?
If the sum of n terms of an A.P. is 2 n2 + 5 n, then its nth term is
If in an A.P., Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
The pth term of an A.P. is a and qth term is b. Prove that the sum of its (p + q) terms is `(p + q)/2[a + b + (a - b)/(p - q)]`.
Show that (x2 + xy + y2), (z2 + xz + x2) and (y2 + yz + z2) are consecutive terms of an A.P., if x, y and z are in A.P.
The sum of n terms of an AP is 3n2 + 5n. The number of term which equals 164 is ______.
