Advertisements
Advertisements
प्रश्न
Find a30 − a20 for the A.P.
−9, −14, −19, −24, ...
उत्तर
In this problem, we are given different A.P. and we need to find `a_30 - a_20`.
A.P. −9, −14, −19, −24, ...
Here
First term (a) = -9
Common difference of the A.P. (d) = - 14 - (-9)
= -14 + 9
= -5
Now as we know
`a_n = a + (n -1)d`
Here we find `a_30` and `a_20`
So for 30 th term
`a_30 = a + (20 - 1)d`
`= -9 + (19)(-5)`
= -9 - 95
= -104
So
`a_30 - a_20 = -154 - (-104)`
=-154 + 104
= -50
Therefore for the given A.P `a_30 - a_20 = -50`
APPEARS IN
संबंधित प्रश्न
For an A.P. t3 = 8 and t4 =12, find the common difference d.
For the following arithmetic progressions write the first term a and the common difference d:
`1/5, 3/5, 5/5, 7/5`
For the following arithmetic progressions write the first term a and the common difference d:
−1.1, −3.1, −5.1, −7.1, ...
Find the common difference and write the next four terms of the following arithmetic progressions:
`-1, (-5)/6, (-2)/3`
The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
Write the expression an- ak for the A.P. a, a + d, a + 2d, ... Hence, find the common difference of the A.P. for which a10 −a5 = 200
Find n if the given value of x is the nth term of the given A.P.
`5 1/2, 11, 16 1/2, 22, ......; x = 550`
In an A.P. a = 4 and d = 0, then find first five terms
Determine k so that k2 + 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP.
If –5, x, 3 are three consecutive terms of an A.P., then the value of x is ______.
