Question

in an A.P. , the first term is 2 and the sum of the first five terms is one fourth of the next five terms, show that 20th term is - 112


Class 11​

Answer 1

Given :-

• First term (a) = 2
• Sum of next five term is $\frac{1}{4}$

Prove :-

• $T_{20}$= - 112

Solution :-

• Sum of 1st five term can be Represent as $S_{5}$

• $\frac{1}{4}$ ( Sum of Next five term) can be written as $\: \: \: \: \: \:\frac{1}{4} (S_{10} - S_{5})$

$\: \: \: \: \: \: \: \rm{S_5 = \frac{1}{4} (S_{10} - S_5)} \\ \: \: \: \: \rm{4 \: S_5} = (S_{10} - S_5) \\ \: \: \: \: \rm{4 \: S_5 +S_5 = S_{10} } \\ \: \: \: \rm{5 \red{ S_5}} = \red{S_{10}}$

Finding Sum where n = 5 , for second one

N= 10 and a = 2 [ Given] for finding Common difference!!

$\: \: \: \: \: \: \: \rm{5 \red{ S_5}} = \red{S_{10}} \\ \small{\rm{5 [ \: \frac{n}{2}(2a + (n - 1)d)]} = [ \: \frac{N}{2} (2a + (N - 1)d \: ]} \\ \small{\rm{5 [ \: \frac{5}{2}(2(2) + (5- 1)d)]} = [ \: \frac{10}{2} ( \: 2(2) + (10 - 1)d \: ]} \\ \implies \bf{ \frac{25}{2} (4 + 4d)} = 5(4 + 9d) \\ \rightarrow \frac{25}{ \cancel2} \times \cancel2 \bf{(2 + 2d) = 20 + 45d} \\ \rightarrow\sf{50 + 50d = 20 + 45d} \\ \rightarrow \sf{50 \: d - 45 \: d} = 20 - 50 \\ \rightarrow \sf{5 \: d = - 30} \\ \rightarrow \sf d = \cancel{\frac { - 30}{5}} \\ \hookrightarrow \boxed{\bf{ \green{d = - 6}}}$

$\: \: \: \: \: \hookrightarrow \bf{T_{20} = a + 19d} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{T_{20 } = 2 + 19( - 6)} \\ \: \: \: \: \: \: \: \sf{T_{20} = 2 - 114} \\ \implies{} \boxed{ \rm{T_{20} = - 112}}$

Hence Proved !!