Question

The difference between 12th and 8th term of an arithmetic sequence is 20. a)write the 30th term of the sequence? b)find the first term of the sequence?
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Answer 1

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

$\rm{\implies Difference\:_{(12th-8th\: terms)}=20}$

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

$\rm{\implies 30th\: terms\:of\:AP=?}$

$\rm{\implies first\: terms\:of\:AP=?}$

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

• By applying formula to calculate the value a and d than find 30th and 1st terms of AP]

$\sf\large\underline{Formula\: used:}$

$\rm{\implies T\:_{(n)}=a+(n-1)d}$

$\rm{\implies T\:_{(12)}=a+(12-1)d}$

$\rm{\implies T\:_{(12)}=a+11d}$

$\rm{\implies T\:_{(8)}=a+(8-1)d}$

$\rm{\implies T\:_{(8)}=a+7d}$

• Given difference here is 20]

$\rm{\implies T_{(12)}-T\:_{(8)}=20}$

$\rm{\implies (a+11d)-(a+7d)=20----(1)}$

$\rm{\implies a+11d-a-7d=20}$

$\rm{\implies 4d=20}$

$\rm{\implies d=5}$

• Now, calculate the value of a here]

$\rm{\implies Let\:1st\:term\:ap\:(a)=1}$

• putting the value of d in eq--(1)and calculate a here]

$\rm{\implies (a+11d)-(a+7d)=20}$

$\rm{\implies a+55-a-35=20}$

$\rm{\implies 20=20}$

$\rm{\implies1=1}$

• By calculate the value of a=1 and d=5
• Now, calculate 30th term here]

$\rm{\implies T\:_{(n)}=a+(n-1)d}$

$\rm{\implies T\:_{(30)}=1+(30-1)5}$

$\rm{\implies T\:_{(30)}=1+29*5}$

$\rm{\implies T\:_{(30)}=1+145}$

$\rm{\implies T\:_{(30)}=146}$

$\sf\large{Hence,}$

$\rm{\implies 30th\:terms\:of\:AP=146}$

$\rm{\implies first\: terms\:of\:AP=1}$