Advertisements
Advertisements
Question
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
Solution
Let a be the first term and d be the common difference of the AP. Then,
as = 0 [ an = a + (n-1) d]
⇒ a + (8-1 ) d = 0
⇒ a+ 7d =0
⇒ a=-7d ..................(1)
Now
`(a_38)/(a_18) = (a + (38-1)d)/(a+ (18-1)d)`
⇒`(a_38)/(a_18) = (-7d + 37 d)/(-7d + 17d ) ` [ From (1)]
⇒`(a_38)/(a_18) = (30d)/(10d) = 3`
⇒` a_38 = 3 xx a_18`
Hence, the 38th term of the AP id triple its 18th term.
APPEARS IN
RELATED QUESTIONS
If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?
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
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.
The sum of the first n terms in an AP is `( (3"n"^2)/2 +(5"n")/2)`. Find the nth term and the 25th term.
If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then \[\frac{S_1}{S_2} =\]
An article can be bought by paying Rs. 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs. 3,000 and every other installment is Rs. 100 less than the previous one, find:
- amount of installments paid in the 9th month.
- total amount paid in the installment scheme.
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`
In an A.P., the sum of first n terms is `n/2 (3n + 5)`. Find the 25th term of the A.P.
