The seventh terms of an a.p. is 12.the tenth terms of the progression is greater than the second term by 8. find the first term and the common difference,and hence,determine which termof the progression first exceeds 400.

Given,
↪T7=12
↪a+(7-1)d=12
↪a+6d=12➡(i)

&,
↪T10=T2+8
↪a+(10-1)d=a+(2-1)d+8
↪a+9d=a+d+8
↪a-a+9d-d=8
↪8d=8
↪d=1➡ANS

▶Put value of d in (i),then,
↪a+6×1=12
↪a=12-6
↪a= 6➡ANS

❇To find the term which exceeds 400,

↪a+(n-1)d=400
↪6+(n-1)×1=400
↪(n-1)×1=394
↪n=394+1
↪n=395 for which the term becomes equal to 400. So n must be greater by 1 than current value of n so that the term becomes greater than 400.

❇Hence, n=396➡ANS
That means, 396th term exceeds the value 400.

❇NOTE:
▶Formula used:
↪Tn=a+(n-1)d
where,
a= 1st term
d=common difference

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000