Question

The 5th term of an AP is 8. The 8th term exceeds three times the 2nd term by 2.
Find the first term, the common difference and the sum of first 15 terms.

Please solve this question.​

Answer 1

GIVEN :-

• $\rm 5^{th}$ term of the AP = 8

• $\rm 8^{th}$ term exceeds three times the $\rm 2^{nd}$ term .

TO FIND :-

• First term of the AP .

• Common difference of the AP .

• Sum of first 15 terms .

SOLUTION :-

 Let the first term be a and the common difference be d .

 According to the first condition ,

   $\longrightarrow \sf a+4d= 8 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(1)$

 According to the second condition ,

 $\longrightarrow\sf a_8 = 3(a_2_+2\\\\\longrightarrow a+7d = 3(a+d)+2\\\\\longrightarrow a+7d = 3a + 3d + 2 \\\\\longrightarrow a-3a+7d-3d=2\\\\\longrightarrow -2a+4d=2\\\\\longrightarrow -a+2d = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(2)$

 

 Add equation (1) and equation (2) ,

 $\longrightarrow \sf 6d = 9 \\\\\longrightarrow d = \dfrac{9}{6}\\\\\longrightarrow d=\dfrac{3}{2}\\\\\longrightarrow\bf d = 1.5$

 

 Substitute d = 1.5 in equation (2) ,

  $\longrightarrow\sf -a +2\times 1.5 = 1\\\\\longrightarrow -a+3 = 1 \\\\\longrightarrow -a = -2 \\\\\longrightarrow \bf a = 2$

  $\sf \boxed {\bf Sum \ of \ first \ n \ terms = \dfrac{n}{2} \times (2a+(n-1)d) }$

 

   Sum of first 15 terms = $\sf \dfrac{15}{2}\times (2\times 2 +(15-1)\times 1.5 )$

                                    = $\sf\dfrac{15}{2}\times (4+14\times 1.5)$

                                    = $\sf\dfrac{15}{2} \times( 4+ 21 )$

                                    =  $\sf \dfrac{15}{2} \times 25$

                                    = 187.5

 

Answer 2

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