Question

if the sum of the 4th term and the 12th term of an A.P is 8 , what is the sum of the first 15 terms of the progression?

Answer 1

Answer: 60

Step-by-step explanation:

a + 3d + a + 11d  = 8

= 2a + 14d = 8

= 15(2a+14d) = 15*8 = 120      (multiplying 15 on both sides)

= 15[2a+(15-1)d]/2 = 120/2 = 60     (dividing 2 on both sides)

= S₁₅ = 60

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Answer 2

nth Term of an Arithmetic Progression, Tn=a+(n−1)dTn=a+(n−1)d

Sum of n terms, Sn=(n/2)∗[2a+(n−1)∗dSn=(n/2)∗[2a+(n−1)∗d

Where, a = first term of Progression, d = common difference (Second term - first term or Third - second term etc.)

4th Term, T4=a+(4−1)∗d=a+3dT4=a+(4−1)∗d=a+3d

12th Term, T12=a+(12−1)∗d=a+11dT12=a+(12−1)∗d=a+11d

Given : (a+11d) + (a+3d) = 8

i.e. 2a + 14d = 8

Question : Sum of 15 terms, S15=(15/2)∗[2a+(15−1)∗d=(15/2)∗[2a+14d]=(15/2)∗8=60S15=(15/2)∗[2a+(15−1)∗d=(15/2)∗[2a+14d]=(15/2)∗8=60

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