Question

Find the sum of the following APs:-

(i) -37,-33,-29,............., to 12th terms.
(ii) 0.6,1.7,2.8,............, to 100 terms.

Hint :- (i) Ans- (-180).
(ii) Ans - 5505.

Solve the sums and get answer given in hint and be a brillient.​ plzz answer fast

Answer 1

Answer:

I) First term, a=-33

Since there are 12 terms so n=12

Difference = a2-a = -33-(-37)=4

Sum of n terms of A.P. is

n/2[2*a +(n-1)*d]

Hence sum of 12 terms of A.P. is

=12/2[(2*-37) + (12–1)*4]

=6[-74+44]

= -180

ii)Sum of n terms of AP formula:

Where First term = a

Common difference = d

Number of terms = n

0.6, 1.7, 2.8,…. to 100 terms.

a = 0.6

d = 1.7 – 0.6 = 1.1

Using (1)

S100 = 100/2 [2(0.6) + (100 – 1)(1.1)] =

(50) × [1.2 + (99 × 1.1) ] = 50 × 110.1 = 5505

Sum of 100 terms of this AP is 5505

hope it helps

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