Advertisements
Advertisements
प्रश्न
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
Advertisements
उत्तर
an = 3 + 4n
a1 = 3 + 4 x 1 = 3 + 4 = 7
a2 = 3 + 4 x 2 = 3 + 8 = 11
a3 = 3 + 4 x 3 = 3 + 12 = 15
a4 = 3 + 4 x 4 = 3 + 16 = 19
and so on Here, a = 1 and d = 11 – 7 = 4
S15 = `n/(2)[2a + (n - 1)d]`
= `(15)/(2)[2 xx 7 + (15 - 1) xx 4]`
= `(15)/(2)[14 + 14 xx 14]`
= `(15)/(2)[14 + 56]`
= `(15)/(2) xx 70`
= 525.
APPEARS IN
संबंधित प्रश्न
Find the sum of all numbers from 50 to 350 which are divisible by 6. Hence find the 15th term of that A.P.
How many terms of the A.P. : 24, 21, 18, ................ must be taken so that their sum is 78?
Show that `(a-b)^2 , (a^2 + b^2 ) and ( a^2+ b^2) ` are in AP.
If the sum of first m terms of an AP is ( 2m2 + 3m) then what is its second term?
Find the sum of n terms of the series \[\left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + . . . . . . . . . .\]
If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
Q.7
Find the sum of odd natural numbers from 1 to 101
Find the sum of the integers between 100 and 200 that are not divisible by 9.
