Advertisements
Advertisements
प्रश्न
The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is `(1)/(3)`. Calculate the first and the thirteenth term.
Advertisements
उत्तर
T10 : T30 = 1 : 3, S6 = 42
Let a be the first term and d be a common difference, then
`(a + 9d)/(a + 29d) = (1)/(3)`
⇒ 3a + 27d = a + 29d
⇒ 3a – a = 29d – 27d
⇒ 2a = 2d
⇒ a = d
Now, S6 = 42
= `n/(2)[2a + (n - 1)d]`
⇒ 42 = `(6)/(2)[2a + (6 - 1)d]`
⇒ 42 = 3[2a + 5d]
⇒ 14 = 2a + 5d
⇒ 14 = 2a + 5a ...(∵ d = a)
⇒ 7a = 14
⇒ a = `(14)/(7)` = 2
∴ a = d = 2
Now, T13 = a + (n – 1)d
= 2 + (13 – 1) x 2
= 2 + 12 x 2
= 2 + 24
= 26
∴ 1st term is 2 and thirteenth term is 26.
APPEARS IN
संबंधित प्रश्न
How many multiples of 4 lie between 10 and 250?
Choose the correct alternative answer for the following question.
For an given A.P. a = 3.5, d = 0, n = 101, then tn = ....
Choose the correct alternative answer for the following question .
15, 10, 5,... In this A.P sum of first 10 terms is...
Fill up the boxes and find out the number of terms in the A.P.
1,3,5,....,149 .
Here a = 1 , d =b`[ ], t_n = 149`
tn = a + (n-1) d
∴ 149 =`[ ] ∴149 = 2n - [ ]`
∴ n =`[ ]`
The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
The sum of first 9 terms of an A.P. is 162. The ratio of its 6th term to its 13th term is 1 : 2. Find the first and 15th term of the A.P.
The given terms are 2k + 1, 3k + 3 and 5k − 1. find AP.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find:
- the production in the first year.
- the production in the 10th year.
- the total production in 7 years.
Determine the sum of first 100 terms of given A.P. 12, 14, 16, 18, 20,......
Activity :- Here, a = 12, d = `square`, n = 100, S100 = ?
Sn = `"n"/2 [square + ("n" - 1)"d"]`
S100 = `square/2 [24 + (100 - 1)"d"]`
= `50(24 + square)`
= `square`
= `square`
Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..
