Advertisements
Advertisements
प्रश्न
The 8th term of an AP is zero. Prove that its 38th term is triple its 18th term.
उत्तर
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
संबंधित प्रश्न
In an AP given an = 4, d = 2, Sn = −14, find n and a.
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?
Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .
If the sum of first p term of an A.P. is ap2 + bp, find its common difference.
Q.3
For an A.P., If t1 = 1 and tn = 149 then find Sn.
Activitry :- Here t1= 1, tn = 149, Sn = ?
Sn = `"n"/2 (square + square)`
= `"n"/2 xx square`
= `square` n, where n = 75
Find the sum of three-digit natural numbers, which are divisible by 4
The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
Sum of 1 to n natural number is 45, then find the value of n.
