Question

in an arithmetic progression a=5 d=10 and the sum of the first n terms is 32,000 find n​

Answer 1

Step-by-step explanation:

using formula

Sₙ = n/2 {2a+(n-1)d}

32000 = n/2{2×5+(n-1)10}

32000×2 = n {10+10n-10}

64000 = 10n²

n² = 6400

n = √6400

n = 80

Answer 2

Answer:

80

Step-by-step explanation:

HEY MATE YOUR ANSWER IS HERE ^_^

GIVEN THAT

a = 5

d = 10

S = 32000

n = ?

$S = \frac{n}{2} [2a+(n-1)d]$

n/2[2*5+(n-1)10] = 32000

n/2[1+n-1] = 3200

$\frac{n^{2}}{2} = 3200\\\\n^{2} = 6400$

n = 80

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