Question

18. what is the value of 21+24+27+......+51?

Answer 1

21+24+27...........+51

IT IS IN A.P

SO

a = 21
d = 3
n = ?
Tn = last term = 51

Tn = a+(n-1)d

51 = 21+(n-1)3

51-21=3n-3

30 = 3n-3

30+3=3n

33=3n

33/3 = n

11= n

Sn =n/2[a+Tn]

S11 = 11/2[21+51]

S11 = 11/2×72

S11 = 11×36

S11 = 396


so the value of 21+24+28. .. .. ....51 is 396

Answer 2

According to the Question

a = 21

l = 51 = Tn

d = 3

${\implies T_{n} = a + (n - 1)d}$

51 = 21 + (n - 1) × 3

Divides both sides by 3 we get

17 = n + 6

n = 17 - 6

n = 11

Now sum :-

${\implies S_{n} = \dfrac{n}{2}[2a + (n - 1)d]}$

${\implies S_{11} = \dfrac{11}{2}[2\times 21 + (11 - 1)3]}$

${\implies S_{11} = \dfrac{11}{2}[42 + 30]}$

${\implies S_{11} = \dfrac{11}{2}\times 72}$

= 11 × 36

= 396