1.

The first and the last terms of an AP are 8 and 350 respectively. If its commondifference is 9, how many terms are there and what is their sum?

Answer»
AP: 8.......350(Given)
A1=8
An=350 
D. =9 
So the AP is, 
A2: a+d = 8+9=17
A3: a+2d = 8+2(9)=27
So we have the AP as 8,17, 26
Now we have to find number of terms present in the AP, 
Using An=a+(n-1)d
350=8+(n-1)9
350-8=(n-1)9
342/9=n-1
38=n-1
n=39
Now, we found out the number of terms, that is 39 terms in total. The next step is calculating the sum of all the 39 terms, so 

Using, Sn=n/2{2a+(n-1)d}
Sn = 39/2{2(8)+(39-1)9}
      =39/2{16+342}
      =39/2*358
      =6981
Therefore, we got the sum as 6981

Hope my answer helps you! 
Thank you.

First term, a=8, Common Difference,  d=9
Last Term, An= 350
a+(n-1)d= 350
8+(n-1)9=350
(n-1)9=350-8=342
(n-1)9= 342
n-1= 342/9= 38
n-1= 38
n= 38+1= 39,
So, there are 39 terms.
Sum of n terms, Sn=n/2(a+An)
S39= 39/2(8+350)
S39= (39/2) * 358
S39= 39* 179= 6981

a=8
An=350
d=9    An=? 

An=a+(n-1) d
350=8+(n-1) 9
350-8=(n-1) 9
342=(n-1) 9
38=n-1
38+1=n
39=n

a=8
d=9
l=350
Sn=? 





Sn=n/2(a+l) 
S39=39/2(8+350) 
S39=39/2(358) 
S39=39(179) 
S39=6981

An= a+(n-1)d
350=8+(n-1)9
350-8=(n-1)9
342=(n-1)9
342/9=n-1
38=n-1
38+1=n

39=n
Sn=n/2[a+an]
Sn=39/2[8+350]
Sn=39/2[358]
Sn=39/2×358
Sn=39×179
Sn=6981



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