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Answer :
Given that,
first term, a = 5
last term, l = 45
Sum of the AP, \(S_n\) = 400
As we know, the sum of AP formula is;
\(\Rightarrow S_n = {{n} \over {2}} (a+l)\)
\(\Rightarrow \) 400 = \({{n} \over {2}}\)(5+45)
\(\Rightarrow \) 400 = \({{n} \over {2}}\)(50)
Number of terms, n =16
As we know, the last term of AP series can be written as;
l = a+(n -1)d
\(\Rightarrow \) 45 = 5 +(16 -1)d
\(\Rightarrow \) 40 = 15d
\(\Rightarrow \) Common difference, d = \({{40} \over {15}} = {{8} \over {3}}\)