Math, asked by hidayatkhan, 1 month ago

The first term of an arithmetic sequence is 6 and the common difference is 4. How many terms must be added together until the sum of the terms is equal to 510?

Answers

Answered by kartik2507
1

Answer:

n = 15

Step-by-step explanation:

Sn = n/2(2a + (n-1)d)

given that

a = 6 d = 4 Sn = 510

510 =  \frac{n}{2} (2(6) + (n - 1)(4)) \\ 510 =  \frac{n}{2} (12 + 4n - 4) \\ 510 =  \frac{n}{2} (4n + 8) \\ 510 =  \frac{n}{2}  \times 2(2n + 4) \\ 510 = n(2n + 4) \\ 510 =  {2n}^{2}  + 4n \\ 510 = 2( {n}^{2}  + 2n) \\  \frac{510}{2}  =  {n}^{2}  + 2n \\ 255 =  {n}^{2}  + 2n \\  {n}^{2}  + 2n - 255 = 0 \\  {n}^{2}  + 17n - 15n - 255 = 0 \\ n(n + 17) - 15(n + 17) = 0 \\ (n + 17)(n - 15) = 0 \\ n + 17 = 0 \:  \:  \:  \:  \:  \: n - 15 = 0 \\ n =  - 17 \:  \:  \:  \:  \:  \: n = 15

we take the positive value of n = 15

therefore the number of terms of the AP to get sum of 510 is 15

verification

 =  \frac{15}{2} (2(6) + (15 - 1)(4)) \\  =  \frac{15}{2} (12 + 14(4)) \\  =  \frac{15}{2} (12 + 56) \\  =  \frac{15}{2}  \times 68 \\  = 15 \times 34 \\  = 510

Hope you get your answer

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