Advertisements
Advertisements
प्रश्न
The 25th term of an A.P. exceeds its 9th term by 16. Find its common difference.
उत्तर
nth term of an A.P. is given by tn= a + (n – 1) d.
`\implies` t25 = a + (25 – 1)d = a + 24d
And t9 = a + (9 – 1)d = a + 8d
According to the condition in the question, we get
t25 = t9 + 16
`\implies` a + 24d = a + 8d + 16
`\implies` 16d = 16
`\implies` d = 1
APPEARS IN
संबंधित प्रश्न
The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts in 4623.
An A.P. consists of 60 terms, If the first and the last terms be 7 and 125 respectively, find the 31st term.
For an A.P., show that (m + n)th term + (m – n)th term = 2 × mthterm
The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31st term of this A.P.
Find the sum of last 8 terms of the A.P. –12, –10, –8, ……, 58.
Find the A.P. whose nth term is 7 – 3K. Also find the 20th term.
Find the nth term and the 12th term of the list of numbers: 5, 2, – 1, – 4, …
How many two digit numbers are divisible by 3?
Justify whether it is true to say that the following are the nth terms of an A.P. 2n – 3
