Question

qusetion number 21 please do it as quick as possible
only answer if u know :)​

Answer 1

ATwo digit odd positive numbers are 11,13,15,17...........99 are in A.P.

Here a = 11 and d = 2, tn= 99, n = ?

Sum of the n terms = (n/2)[2a+(n -1)d]

But tn = a + (n -1)d

⇒ 99 = 11+ (n-1)2

⇒ 99 -11 = (n-1)2

⇒ 88/2 = (n-1)

∴ n = 44

thanku

Answer 2

the formula we have to use here is

Sn = n/2(a + L) where a is the first digit and L is the last digit

here ; the odd numbers that are 2 digits are starting from 11 , so the AP hence formed is 11 , 13, 15 ..... 99

here , a = 11 d =2 and L=An = 99 (An and L are the same , change the variable according the formula )

so to find n ->

An = a + (n-1)d
99= 11 + (n-1) 2
99= 11 + 2n - 2
99= 9 + 2n
99-9= 2n
90 = 2n
90/2 = n therefore n= 45


now use it in the first equation

Sn = 45/2( 11 + 99)

=45/2 × 110
=45 × 55
= 2475


this is your answer , please mark brainlist