Advertisements
Advertisements
प्रश्न
If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 : 3n + 4, then write the ratio of their m th terms.
Advertisements
उत्तर
Given:
\[\frac{S_n}{{S_n}^1} = \frac{2n + 5}{3n + 4}\]
\[ \Rightarrow \frac{\frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}}{\frac{n}{2}\left\{ 2b + \left( n - 1 \right) d^1 \right\}} = \frac{2n + 5}{3n + 4}\]
\[ \Rightarrow \frac{2a + \left( n - 1 \right)d}{2b + \left( n - 1 \right) d^1} = \frac{2n + 5}{3n + 4} . . . \left( 1 \right)\]
Ratio of their m terms =\[\frac{a_m}{b_m}\]
To find the ratio of the mth terms, replace n by 2m \[-\] 1 in equation (1).
\[\Rightarrow \frac{2a + \left( 2m - 2 \right)d}{2b + \left( 2m - 2 \right) d^1} = \frac{2\left( 2m - 1 \right) + 5}{3\left( 2m - 1 \right) + 4}\]
\[ \Rightarrow \frac{a + \left( m - 1 \right)d}{b + \left( m - 1 \right) d^1} = \frac{4m - 2 + 3}{6m - 3 + 4}\]
\[ \Rightarrow \frac{a_m}{b_m} = \frac{4m + 1}{6m + 1}\]
APPEARS IN
संबंधित प्रश्न
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Find the sum to n terms of the A.P., whose kth term is 5k + 1.
Find the sum of all numbers between 200 and 400 which are divisible by 7.
Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
If the nth term an of a sequence is given by an = n2 − n + 1, write down its first five terms.
Which term of the A.P. 4, 9, 14, ... is 254?
Which term of the sequence 24, \[23\frac{1}{4,} 22\frac{1}{2,} 21\frac{3}{4}\]....... is the first negative term?
The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.
Find the 12th term from the following arithmetic progression:
3, 8, 13, ..., 253
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term.
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.
Find the sum of the following arithmetic progression :
1, 3, 5, 7, ... to 12 terms
Find the sum of the following arithmetic progression :
41, 36, 31, ... to 12 terms
Find the sum of the following arithmetic progression :
\[\frac{x - y}{x + y}, \frac{3x - 2y}{x + y}, \frac{5x - 3y}{x + y}\], ... to n terms.
Find the sum of the following serie:
(a − b)2 + (a2 + b2) + (a + b)2 + ... + [(a + b)2 + 6ab]
Find the sum of all natural numbers between 1 and 100, which are divisible by 2 or 5.
The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by \[10 \frac{1}{2}\] , find the number of terms and the series.
If 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?
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).
If \[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P., prove that:
\[\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}\] are in A.P.
If a, b, c is in A.P., prove that:
(a − c)2 = 4 (a − b) (b − c)
We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
Write the sum of first n even natural numbers.
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
If the sum of n terms of an A.P., is 3 n2 + 5 n then which of its terms is 164?
If n arithmetic means are inserted between 1 and 31 such that the ratio of the first mean and nth mean is 3 : 29, then the value of n is
Mark the correct alternative in the following question:
The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is
Mark the correct alternative in the following question:
Let Sn denote the sum of first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to
If a1, a2, ..., an are in A.P. with common difference d (where d ≠ 0); then the sum of the series sin d (cosec a1 cosec a2 + cosec a2 cosec a3 + ...+ cosec an–1 cosec an) is equal to cot a1 – cot an
If the sum of p terms of an A.P. is q and the sum of q terms is p, show that the sum of p + q terms is – (p + q). Also, find the sum of first p – q terms (p > q).
The sum of n terms of an AP is 3n2 + 5n. The number of term which equals 164 is ______.
