Question

explain its clearly and don't scam​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Your question needs a correction.

Correct question : - how many terms of the AP 2, 4, 6, 8, 10 are needed to get sum 210.

Solution : -

Let there are n terms.

Given, 1st term = a = 2

2nd term = a₂ = 4

3rd term = a₃ = 6

∴ Common Difference = a₃ - a₂

= 6 - 4

= 2

we know that the sum of n terms of AS is \dfrac{n}{2}{2a+(n-1)d] .

Substituting the given values in the formula given above.

210 =n/2 [2(2) + (n − 1)(2)

210 = n/2 [4 + 2n - 2]

210 = n/2 [2n + 2]

210 = n/2 (2(n+1)]

4G TA 46

210 = n(n+1)

210 = n^2 + n

210 = n( n + 1 )

210 = n^2 + n

n^2 + n - 210 = 0

n^2 + ( 15 - 14 )n - 210 = 0

n^2 + 15n - 14n - 210 = 0

n( n + 15 ) - 14( n + 15 ) = 0

( n + 15 ) ( n - 14 )= 0

n = - 15 or n = 14

As n is the number of terms of AP, it can't be negative. Therefore, n = 14