Question

SOLVE 16 one plzzzzzz

Answer 1

In this first term is 5 and 10 is the common difference

Therefore 31 term is

a+29d=5+290=295

Now 130 more then 31 term so

a+nd=295+130

n is the required term

5+n(10)=425

10n=420

n=42

So answer is 42 term

And hope it helps

Answer 2

HERE IS YOUR ANSWER..⬇⬇

$\underline{SOLUTION}$ ➡

Given,

10th term, a10 = 52
                          
=> a + +(10-1)d = 52

=> a + 9d = 52 ...........(1)

Also,  

  a17 = 20+a13   

=> a+(17-1)d = 20 + {a+ (13-1)d}
    
  => a + 16d = 20+(a+12d)
      
=> a + 16d = 20 + a + 12d

=> a - a + 16d - 12d = 20

=> 16d-12d = 20 
         
=> 4d = 20    

=> d = 20/4

=> d = 5    

Putting the value of "d" in eq. (1),

we get, 
            
a + 9×(5) = 52   
        
=> a +45 = 52     
         
=> a = 52 - 45

=> a = 7    

Hence, the required AP is 7,12,17,22,27,.........

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