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.
You need to login to perform this action.
You will be redirected in
3 sec