Question

SOLVE 16 one plzzzzzz

Answer 1

let a be the first term and d be the common difference
we know that
Tn=a+(n-1)d
52=a+9d
T17=a+16d
T13=a+12d
a.t.q
T17=T13+20
a+16d=a+12d+20
a and a will be cancel out
16d-12d=20
4d=20
d=20/4
d=5
on putting d=5 in above equation
52=a+9d
52=a+45
a=52-45
a=7
a.p will be
7,12,17,22.......
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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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