Question

The sixth term of an arithmetic sequence is 14. The sum of the second and fourth terms is −2 Determine the first three terms of the sequence Select one: a. −11;−16;−21 b. −11;−6;−1 c. −16;−11;−6

Answer 1

$\sf\large\underline{Given:}$

$\tt{\implies 6th\: term\:of\:AP=14}$

$\tt{\implies sum\:of\:(2nd+4th)\: term\:of\:AP=-2}$

$\sf\large\underline{To\: Find:}$

• First 3 term of AP=?]

$\sf\large\underline{Solution:}$

• By applying formula here to calculate sum]
• Tn=a+(n-1)d]

$\tt{\implies Sum\:_{(2nd+4th)}=-2}$

$\tt{\implies a+(2-1)d+a+(4-1)d=-2}$

$\tt{\implies a+d+a+3d=-2}$

$\tt{\implies 2a+4d=-2-----(1)}$

• Calculate the 6th term AP]

$\tt{\implies a+(6-1)d=14}$

$\tt{\implies a+5d=14-----(2)}$

• In eq 2 multiplying by 2 than subtracting from eq 1]

$\tt{\implies 2a+10d=28}$

$\tt{\implies 2a+4d=-2}$

• by solving We get here]

$\tt{\implies 6d=30}$

$\tt{\implies d=5}$

• Putting the value of d=5 in eq 1]

$\tt{\implies 2a+4d=-2}$

$\tt{\implies 2a+4(5)=-2}$

$\tt{\implies 2a+20=-2}$

$\tt{\implies 2a=-22}$

$\tt{\implies a=-11}$

• now calculate 1st 3 term of sequence here]

=>1st term=a-d=-11-5=-16

=>2nd term=a=-11

=>3rd term=a+d=-11+5=-6

$\sf\large{Hence,}$

$\rm{\implies The\: sequence\:are\:-16,\:-11\:,-6}$

$\tt{\underbrace{Answer-option(c)}}$