Question

Find the sum of first 20 terms of an A.P., in which 3rd term is 8 and 7th term is 9 more than its 4th term. ​

Answer 1

Answer:

$S{__{20}} = 197.5$

Step-by-step explanation:

The equation for A.P. is $a{__n} = a{__1}+(n-1)d$

where An is the value of the nth sequence,

A1 is the value of the first sequence,

n is the number of sequence,

and d is the difference between each term in the A.P.

Given:

$a{__1} + 2d = 8$

$a{__1} + 6d = 9$

We can solve two unknowns with simulataneous equations.

$a{__1} = a = \frac{15}{2}$, $d = \frac{1}{4}$

Now use the equation for sum of A.P., that is $S{__n} = \frac{n}{2}(2a+(n-1)d)$.

Plug in all the values.

$S{__{20}} = \frac{20}{2}(2*(\frac{15}{2}) +(20-1)(\frac{1}{4} ))$

$S{__{20}} = 197.5$

Answer 2

Answer:

Given That:-

• 3rd term of A.P. is 8.
• 7th term is 9 more than it's 4th term.

$\:$

We have:

$\dashrightarrow{ \green{ \rm{a_3 = 8}}}$

${ \dashrightarrow{ \green{ \rm {\: a_7 = a_4 + 9}}}}$

To Find:

• Sum of 20 term of an A.P.

$\:$

So, we know:

• a + 2d = 8 ...(1)
• a + 6d = a + 3d + 9 ...(2)

$\:$

$\star \: { \boxed{ \pink{ \rm { a_3 = 8}}}} \: \star$

${ \large{ \dashrightarrow{{ \rm{a \: + \: 6d \: = \: a \: + \: 3d \: + \: 9 \: }}}}}$

$\:$

${ \large{ \dashrightarrow{ \red{ \rm{a \: + \: 6d \: = \: a \: + \: 3d \: + \: 9}}}}}$

${ \large{ \dashrightarrow{ \red{ \rm{6d \: - 3d \: = \: 9}}}}}$

${ \large{ \dashrightarrow{ \red{ \rm{3d \: = \: 9}}}}}$

${ \large{ \dashrightarrow{ \red{ \rm{d \: = \: \frac{9}{3} }}}}}$

${ \large{ \dashrightarrow{ \red{ \rm{d \: = \: { \green{ \boxed{ \rm{3}}}}}}}}}$

So, putting the value of 'd' in Equation (1), we get:

${ \large{ \longrightarrow{ \rm{a \: + \: 2(3) \: = \: 8}}}}$

${ \large{ \longrightarrow{ \rm{a \: + \: 6 \: = \: 8}}}}$

${ \large{ \longrightarrow{ \rm{a \: = \: 8 \: - \: 6}}}}$

${ \large{ \longrightarrow{ \rm{ a\: = \:{ \green {\boxed{ 2}}}}}}}$

Now, Sum of 20th term of A.P. is :

${ \large{ \boxed{ \blue{ \rm{S_n \: = a + \frac{n}{2} \left[2a + (n - 1)d \right]}}}}}$

So, Putting:

• a = 2
• d = 3
• n = 20

${ \longrightarrow{ \rm{\large{S_n} \: = \: a \: + \frac{n}{2} \left[ 2a \: + (n - 1)d\right]}}}$

${ \longrightarrow{ \rm{ \large{S_n \: = \: 2 + \frac{20}{2} \left[ \: 2(2) + (20 - 1)(3) \: \right]}}}}$

${ \longrightarrow{ \rm{ \large{ S_n \: = \: 2 + 10\left[ 4 + (19)(3)\right]}}}}$

${ \longrightarrow{ \rm{ \large{S_n \: = \: 2 + 10\left[4 + 57 \right]}}}}$

${ \longrightarrow{ \rm{ \large{S_n \: = \: 2 + 10 \: \left[ \: 61 \: \right]}}}}$

${ \longrightarrow{ \rm{ \large{S_n \: = \: 2 + 610}}}}$

${ \longrightarrow{ \large{ \rm{S_n \: = \: { \green{ \boxed{612}}}}}}}$

Hence,

${ \boxed{ \pink{ \rm{Sum \: of \: 20 ^{th} \: term \: = \: 612 \: }}}}$