Question

in an AP first term is 2. the sum of first five term is one fourth of the sum of next 5 terms. find the sum of first 20 term​

Answer 1

${ \bold {\underline{ \underline{Arithmetic \: Progressions}}}}$

Given :

•First term (a) → 2

• Sum of first 5 terms = ¼[ Sum of next five terms ]

Using the formula for Sum of n terms :

$= > S_n = \frac{n}{2} [2a + (n - 1)d \: ]$

For Sum of first five terms ,

$= > \: S_5 = \frac{5}{2} [ \: 2 \times 2 + (5 - 1)d \: ] \\ \\ = > \: S_5 = \frac{5}{2} \: \: [ 4 + 4d] \\ \\ = > \: S_5 = \frac{5}{2} \times 4 \: \: [ 1+ d] \\ \\ { \bold{S_5 = 10[1 + d]}}$

For the Sum of first 10 terms ,

$= > S10 = \frac{10}{2} [ \: 2 \times 2 + (10 - 1)d \: ] \\ \\ = >{ \bold { S10 = 5 [ 4 + 9d \ ]}}$

Now , as per question :

$\: S_5 = \frac{1}{4} \: [ \: S10 - \: S_5 \: ] \\ \\ = > 10 + 10d = \frac{1}{4} \: [ \: 20 + 45d \: - 10 - 10d \: ] \\ \\ = > 40 + 40d = 10 + 35d \\ \\ = > 5d = - 30 \\ \\ = > { \boxed{ \bold {d = - 6}}}$

•°• Sum of first 20 terms :-

$= > \: S20 = \frac{20}{2} \: [ \: 2 \times 2 + (20 - 1) - 6 \: ] \\ \\ = > \: S20 = 10 \: (4 + 19 \times ( - 6))$

=> S20 = 10 [ 4 - 114 ]

=> S20 = 10 × (-110)

•°• ${\underline{\bold{S20 = -1100}}}$

Answer 2

☞ In an AP first term is 2.

=> a (first term) = 2

☞ The sum of first five term is one fourth of the sum of next 5 terms.

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

• A.T.Q.

Sum of mean of first 5 terms = Sum of first 10 terms - Sum of first 5 terms

=> $S_{10}$ - $S_{5}$

=> $\dfrac{10}{2}$ [2(2) + (10 - 1)d] - $\dfrac{5}{2}$ [2(2) + (5 - 1)d]

=> 5 (4 + 9d) - $\dfrac{5}{2}$ (4 + 4d)

=> 20 + 45d - $\dfrac{5}{2}$ × 4 (1 + d)

=> 20 + 45d - 10 (1 + d)

=> 20 + 45d - 10 - 10d

=> 45d - 10d + 20 - 10

=> 35d + 10

_______________________________

Sum of mean of first 5 terms = $\dfrac{1}{4}$ × Sum of next 5 terms

=> $\dfrac{5}{2}$ (4 + 4d) = $\dfrac{1}{4}$ × (10 + 35d)

=> $\dfrac{5}{2}$ × 4 (1 + d) = $\dfrac{1}{4}$ × (10 + 35d)

=> 10(1 + d) = $\dfrac{1}{4}$ × (10 + 35d)

=> 10 + 10d = $\dfrac{1}{4}$ × (10 + 35d)

=> 4(10 + 10d) = (10 + 35d)

=> 40 + 40d = 10 + 35d

=> 40 - 10 = 35d - 40d

=> 30 = - 5d

=> d = - 6

______________________________

☞ We have to find sum of first 20 terms.

$S_{n}$ = $\dfrac{n}{2}$ [2a + (n - 1)d]

n = 20

a = 2

d = - 6

=> $S_{20}$ = $\dfrac{20}{2}$ [2(2) + (20 - 1)(-6)]

=> $S_{20}$ = 10[4 + 19(-6)]

=> $S_{20}$ = 10(4 - 114)

=> $S_{20}$ = 10 (-110)

=> $S_{20}$ = - 1100

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Sum of first 20 terms is - 1100

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