Find the sum of the arithmetic sequence. 3, 5, 7, 9, 21

Given the arithmetic sequence

3, 5, 7, 9, ..., 21

The first term a = 3 & a common difference d = 5 - 3 = 7 - 5 = ... = 2

If there are n number of terms in the above A.P. then last term l = 21 will be the nth term given as:

l = a + (n - 1)d

21 = 3 + (n − 1)2

n = 10

Hence the sum of given arithmetic progression (A. P.) up to 10 terms is given general formula.

Sn = n/2(a + l)

S10 = 10/2(3 + 21)

= 120

secant-secant formula:

(whole secant) x (external part) =           (whole secant) x (external part)

(8+y-2) *8=                                           (y+4+6) *6

(y+6)*8 = (y+10) *6

distribute

8y+48 = 6y +60

subtract 6y from each side

2y +48 = 60

subtract 48 from each side

2y = 12

divide by 2

y = 6

sv =     (y+4+6)

sv   = 6+4+6

sv = 16