Question

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Q.The sum of first $q$ terms of an AP is $(63{q}^{} - 3{q}^{2})$. If its $p$th term is -60, find the value of $p$ . Also, find the 11th term of its AP.​ ​

Answer 1

$\huge\dag$ $\huge\fcolorbox{red}{yellow}{Answer}$$\huge\dag$

pth term =ap = a + d(p−1) =q

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1)

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1)

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we get

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we getap+q=0

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we getap+q=0 [A straightforward use of an=a+(n−1)d ]

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we getap+q=0 [A straightforward use of an=a+(n−1)d ]QED

Hope this helps you mate

Answer 2

Answer:

\huge\dag† \huge\fcolorbox{red}{yellow}{Answer}Answer \huge\dag†

pth term =ap = a + d(p−1) =q

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1)

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1)

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we get

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we getap+q=0

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we getap+q=0 [A straightforward use of an=a+(n−1)d ]

pth term =ap = a + d(p−1) =q Similarly aq=a+d(q−1)=p Then, equate a from both equations, and you'll obtain d=−1 .Replacing in anyone of the above equation, we get, a=p+q−1 .Now,ap+q=a+d(p+q−1) =a−(p+q−1) Using what we proved above, we getap+q=0 [A straightforward use of an=a+(n−1)d ]QED

Hope this helps you mate