question_answer

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is\[7:15\]. Find the numbers.

Answer:

Let the first term of AP be a and d be the common difference.

Let four consecutive terms of an AP be \[a-3d,a-d,a+d\] and \[a+3d\]

According to the question,

\[a-3d+a-d+a+d+a+3d=32\]

\[\Rightarrow \]                                      \[4a=32\]

\[\Rightarrow \]                                         \[a=8\]                                                ?(i)

Also

\[(a-3d\text{ (}a+3d):(a-d)\text{ (}a+d)=7:15\]

\[\frac{{{a}^{2}}-9{{d}^{2}}}{{{a}^{2}}-{{d}^{2}}}=\frac{7}{15}\]

\[\frac{64-9{{d}^{2}}}{64-{{d}^{2}}}=\frac{7}{15}\]

[From (i) put a = 8]

\[15(64-9{{d}^{2}}\text{)}=7(64-{{d}^{2}})\]

\[960-135{{d}^{2}}=448-7{{d}^{2}}\]

\[960-448=135{{d}^{2}}-7{{d}^{2}}\]

\[512=128{{d}^{2}}\]

\[{{d}^{2}}=\frac{215}{128}\]

\[{{d}^{2}}=4\]

\[\Rightarrow \]                      \[d=\pm 2\]

For \[d=2\], four terms of AP are,

\[a-3d=8-3(2)=2\]

\[a-d=8-2=6\]

\[a+d=8+2=10\]

\[a+3d=8+3(2)=14\]

For \[d=-~2\], four terms are

\[a-3d=8-3(-2)=14\]

\[a-d=8-(-2)=10\]

\[a+d=8+(-2)=6\]

\[a+3d=8+3(-2)=2\]

Thus, the four terms of AP series are 1, 6, 10, 14 or 14, 10, 6, 2.