Sum of 1
st and 31st terms of an arithmetic sequence is 80.
a) What is the sum of 2nd and 30th terms?
b) What is the sum of 5th and 27
th terms?
c) What is the 16
th term?
d) What is the sum of first 31 terms?
Answers
Answer:
The sum of 1st term and 31st term of the A.P is 50.
Step-by-step explanation:
Let the first term be 'a' and common difference be 'd'.
nth term of an A.P is given by,
Sum of 2nd and 30th term is 50.
a + (2 – 1)d + a + (30 – 1)d = 50
a + d + a + 29d = 50
2a + 30d = 50
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We've to find the sum of 1st and 31st terms.
= a + a + (31 – 1)d
= a + a + 30d
= 2a + 30d
= 50
Therefore, the sum of 1st term and 31st term of the A.P is 50.
Answer:
Step-by-step explanation:
a1 + a31= 80 (given)
a) a2 + a30 = 80
(since, a2 = a1 + d; a30 = a31 - d;
so, a2 + a30 = a1 + d + a31 - d = a1 + a31 = 80)
b) Similarly, a5 + a27 = 80 ( same explanation as above)
i.e., a5 + a27 = (a1 + 4d) + a31-4d) = a1 + a31 = 80
c) Similarly, a15 + a17= 80
The middle term, a16 will be half the sum of a15 & a17
i.e., a16 = 80/2 = 40
d) Also Note: Sum of a A.P. sequence = middle term * no:of terms
Here, a16 is the middle terms of the entire sequence
so, S31 = a16 * 31 = 40 * 31 -= 1240