Answer:
In the given AP, let first term \[=a\] and common difference \[=d\] Then, \[{{T}_{n}}=a+(n-1)d\] \[\Rightarrow {{T}_{14}}=a+(14-1)d=a+13d\] and \[{{T}_{8}}=a+(8-1)d=a+7d\] Now, \[{{T}_{14}}=2{{T}_{8}}\] (Given) \[a+13d=2(a+7d)\] \[a+13d=2a+14d\] \[a=-d\] ?(i) Also, \[{{T}_{6}}=a+(6-1)d\] \[\Rightarrow a+5d=-8\] ?(ii) Putting the value of a from eq. (i), we get \[-d+5d=-8\] \[4d=-8\] \[d=-2\] Substituting \[d=-2\] in eq. (ii), we get \[a+5(-2)=-8\] \[a=10-8\] \[a=2\] \[\therefore \] Sum of first 20 terms is \[{{S}_{20}}=\frac{n}{2}[2a+(n-1)d]\] \[=\frac{20}{2}[2\times 2+(20-1)(-2)]\] \[=10[4-38]\] \[=-340\]
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