Find the sum of 6-514-412- to 21 terms-

S=6+514+412+... to 21 terms. a=6, d=214 6=21 244= 34, n=21 ∴ S_21=212[2× 6+20× 34] ∴ S_21=212[12 15]= 632= 3112 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Collinearity Condition

Find the sum ...

Question

Find the sum of $6 + 5 \frac{1}{4} + 4 \frac{1}{2} + . . .$ to $21$ terms.

Open in App

Solution

Suggest Corrections

Similar questions

\frac{3}{1 \cdot 2} (\frac{1}{2}) + \frac{4}{2 \cdot 3} {(\frac{1}{2})}^{2} + \frac{5}{3 \cdot 4} {(\frac{1}{2})}^{3} + \dots

20

38 + 35 + 32 + . . . . . + 2

6 + 5 \frac{1}{4} + 4 \frac{1}{2} + . . . . 25 t e r m s

{[{(4)}^{\frac{1}{4}} \times {(2)}^{\frac{1}{2}} \times {(5)}^{\frac{1}{5}}]}^{0}

Find the sum of the series s = 1 + $\frac{1}{2}$ (1 + 2) + $\frac{1}{3}$ (1 + 2 + 3) + $\frac{1}{4}$ (1 + 2 + 3 + 4) + .........upto 40 terms.

Join BYJU'S Learning Program