If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1:7, and a+n=33, then the value of n is:
Solutions · 1
Official solution
a, A1, A2 ........... An, 100
Let d be the common difference of above A.P. then
\[{{a + d} \over {100 - d}} = {1 \over 7}\]
\[ \Rightarrow 7a + 8d = 100\] ...... (i)
and a+n=33 ..... (ii)
and 100=a+(n+1)d
\[ \Rightarrow 100 = a + (34 - a){{(100 - 7a)} \over 8}\]
\[ \Rightarrow 800 = 8a + 7{a^2} - 338a + 3400\]
\[ \Rightarrow 7{a^2} - 330a + 2600 = 0\]
\[ \Rightarrow a = 10,\,{{260} \over 7},\] but \[a \ne {{260} \over 7}\]
\[\therefore\] n=23