What is the sum of the first 25 terms of an arithmetic sequence if the sum of its 8th and 18th term is 72?
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The sum of the first n terms of an arithmetic sequence Sn = \n \over 2 \\) \\ \left\lbrack 2a_1 + \lbrack n-1 \rbrack d \right\rbrack \\)
Where a1 is the first term, 'n' is the number of terms and 'd' is the common difference.
To find the sum of the first 25 terms of the arithmetic sequence, you need the first term t1 and the common difference 'd'.
Data given: Sum of 8th term and 18th term of the sequence is 72.
The 8th term of the arithmetic progression, t8 = t1 + (8 - 1)d
or t8 = t1 + 7d
The 18th term of the sequence t18 = t1 + (18 - 1)d
or t18 t1 + 17d
Sum of the 8th and 18th term is 72
t8 + t18 = t1 + 7d + t1 + 17d = 72
or 2(t1) + 24d = 72 .... (eqn 1)
The sum of the first n terms of an arithmetic sequence Sn = \n \over 2 \\) \\ \left\lbrack 2t_1 + \lbrack n-1 \rbrack d \right\rbrack \\)
The sum of the first 25 terms of an arithmetic sequence S25 = \25 \over 2 \\) \\ \left\lbrack 2t_1 + \lbrack 25-1 \rbrack d \right\rbrack \\)
S25 = \25 \over 2 \\) \\ \left\lbrack 2t_1 + 24d \right\rbrack \\) ... (eqn 2)
The missing information is the value of \\ \left\lbrack 2t_1 + 24d \right\rbrack \\)
In Step 1 in eqn (1), we found the value of \\ \left\lbrack 2t_1 + 24d \right\rbrack \\) as 72
Substituting the value of 2t1 + 24d as 72 in eqn (2), we get S25 = \25 \over 2 \\) \\ \left\lbrack 72 \right\rbrack \\)
S25 = 25 * 36 = 900
Therefore, sum of the first 25 terms S25 = 25 * t13.
We will have the answer when we compute the value of t13
In an AP, the value of any term of the sequence is the arithmetic mean of two terms equidistant from it.
For instance, t2 = \t_1 + t_3 \over 2 \\) and t5 = \t_2 + t_8 \over 2 \\)
So, t13 = \t_8 + t_{18} \over 2 \\)
We know that t8 + t18 = 72
Hence, t13 = \t_8 + t_{18} \over 2 \\) = \72 \over 2 \\) = 36
Sum of the first 25 terms, S25 = 25 * 36
S25 = 900
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