Advertisements
Advertisements
प्रश्न
Choose the correct alternative answer for the following question .
If for any A.P. d = 5 then t18 – t13 = ....
विकल्प
5
20
25
30
Advertisements
उत्तर
It is given that,
d = 5
Now,
\[t_n = a + \left( n - 1 \right)d\]
\[ t_{18} - t_{13} = \left( a + \left( 18 - 1 \right)d \right) - \left( a + \left( 13 - 1 \right)d \right)\]
\[ = \left( a + 17d \right) - \left( a + 12d \right)\]
\[ = 5d\]
\[ = 5\left( 5 \right)\]
\[ = 25\]
APPEARS IN
संबंधित प्रश्न
The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money does the contractor have to pay as a penalty if he has delayed the work by 30 days.
If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero
Find the sum of all odd natural numbers less than 50.
If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.
Write the next term for the AP` sqrt( 8), sqrt(18), sqrt(32),.........`
Find the first term and common difference for the A.P.
127, 135, 143, 151,...
The 19th term of an A.P. is equal to three times its sixth term. If its 9th term is 19, find the A.P.
If Sr denotes the sum of the first r terms of an A.P. Then , S3n: (S2n − Sn) is
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
Q.11
In an A.P., the sum of its first n terms is 6n – n². Find is 25th term.
Find the sum of three-digit natural numbers, which are divisible by 4
Find the next 4 terms of the sequence `1/6, 1/4, 1/3`. Also find Sn.
Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4.
The first term of an AP of consecutive integers is p2 + 1. The sum of 2p + 1 terms of this AP is ______.
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?
Solve the equation
– 4 + (–1) + 2 + ... + x = 437
