The common difference of an arithmetic sequence is 6 and it's one term is 25.Can the sum of any 15terms in this sequence be 2018?why?
Answers
Answered by
1
Step-by-step explanation:
Sum of n terms in this A.P. is
S=n/2(2a+(n-1)d)=2018
15/2(2a+(15–1)6)=2018
2a=((2018×2)/15)–84
a=2776/30
Answered by
0
Answer:
yes
Step-by-step explanation:
Solution: S15 = 2018 = (15/2)[2a +14*6]
2018 = 15[a+42]
2018 = 15a + 630, or
15a = 2018–630 = 1388
a = 1388/15 = 92.933
The AP is 92.533, 98.533, 104.533 …. for 15 terms. Such an AP is possible. Answer.
Check : S15 = (15/2)[2*92.533 + 14*6]
= (15/2)[185.07+84]
= 7.5* 269.07
= 2018. Correct.
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