Question

the sum of the first six term of an arithmetic series is 225 and that of the first nine terms is 315. find the first four termd of the series​

Answer 1

Answer:

Step-by-step explanation:

ANSWER:-

• The sum of first 6 terms is 225
• The sum of first 9 terms is 315.

So, we need to find the series.

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

• S(n) is the sum of numbers
• n is the number of terms
• a is the first number
• d is the common difference.

For first 6 terms:-

$\sf{225 = \dfrac{6}{2}[2a+5d]}$

$\bf{225 = 6a+30d}$

2a+5d = 75 (Taking 3 common) ___________________(2)

For first 9 terms:-

$\sf{315 = \dfrac{9}{2}[2a+8d]}$

$\bf{315 = 9a+36d}$

2a+8d = 70 _____________(1)

Subtracting (2) from (1), we get,

5a-8d = 5

-3d = 5

d = $\dfrac{-5}{3}$

Placing it in (2),

75 = 2a $\dfrac{-25}{3}$

a = 125/3

So second , third and fourth term is:-

$\dfrac{120}{3}$=40

$\dfrac{115}{3}$

$\dfrac{110}{3}$

Things to Note:-

• Remember the formula given above
• $\boxed{\sf{S_n = \dfrac{n}{2}[2a+(n-1)d]}}$
• $\boxed{\sf{a_n = a+(n-1)d}}$
• It is also referred as "l".
• a refers as the first term.
• $a_n$ is the last $n^{th}$ term.

Answer 2

$\huge\bold{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• The sum of the 1st six term of an A.P is 225 .

• And the sum of the 1st nine term of an A.P is 315 .

TO FIND :-

• The first four term of the series .

FORMULA :-

✍️ The sequence of an A.P is,

☞ a + (a + d) + (a + 2d) + ..... + [a + (n - 1) d]

✍️ So, the sum of 1st nth term is,

$\checkmark\:\mathcal{\purple{\boxed{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)\:d]\:}}}$

Where,

• n = number of term .

• a = first term .

• d = common difference .

CALCULATION :-

✍️ Let,

• The first term be ‘a’ .

• The common difference be ‘d’ .

✍️ According to the question, the sum of 1st six term of the A.P is ‘225’ .

• Here, n = 6

$\rm{\implies\:S_6\:=\:\dfrac{6}{2}\:[2a\:+\:(6\:-\:1)\:d]\:}$

$\rm{\implies\:225\:=\:3\:[2a\:+\:5d]\:}$

$\rm{\green{\implies\:2a\:+\:5d\:=\:75\:}}$-----(1)

✍️ Again, the sum of the 1st nine term of the A.P is ‘315’ .

$\rm{\implies\:S_9\:=\:\dfrac{9}{2}\:[2a\:+\:(9\:-\:1)\:d]\:}$

$\rm{\implies\:315\:=\:\dfrac{9}{2}\:[2a\:+\:8d]\:}$

$\rm{\implies\:2a\:+\:8d\:=\:315\:\times{\dfrac{2}{9}}\:}$

$\rm{\green{\implies\:2a\:+\:8d\:=\:70\:}}$-----(2)

✍️ Now equation(1) - equation(2), we get

$\rm{\implies\:2a\:+\:5d\:-\:(2a\:+\:8d)\:=\:75\:-\:70\:}$

$\rm{\implies\:2a\:+\:5d\:-\:2a\:-\:8d\:=\:5\:}$

$\rm{\implies\:-3d\:=\:5\:}$

$\rm{\boxed{\implies\:d\:=\:\dfrac{-5}{3}\:}}$

✍️ Putting the value of ‘d = -5/3’ in equation(1), we get

$\rm{\implies\:2a\:+\:5\times{\dfrac{-5}{3}}\:=\:75\:}$

$\rm{\implies\:2a\:=\:75\:+\:\dfrac{25}{3}\:}$

$\rm{\implies\:a\:=\:\dfrac{250}{3}\:\times{\dfrac{1}{2}}\:}$

$\rm{\boxed{\implies\:a\:=\:\dfrac{125}{3}\:}}$

✍️ So, first term is ‘125/3’ .

✍️ Second term = $\rm{\dfrac{125}{3}\:-\:\dfrac{5}{3}\:=\:40\:}$

✍️ Third term = $\rm{\dfrac{125}{3}\:-\:\dfrac{10}{3}\:=\:\dfrac{115}{3}\:}$

✍️ Four term = $\rm{\dfrac{125}{3}\:-\:\dfrac{15}{3}\:=\:\dfrac{110}{3}\:}$

✔️ Hence, the first four term of the series are “$\rm{\dfrac{125}{3}\:,\:40\:,\:\dfrac{115}{3}\:and\:\dfrac{110}{3}\:}$”.