Question

The sum of n terms of an A.P. is: n(3n+37)/2. If its pth term is 56, then what is the value of p?
a) 10 b) 12 c) 13 d) 15​

Answer 1

Given :

• S{n} = n(3n+37)/2
• a{p} = 56

To find : Value of p

Formula used :

• a{n} = a + (n-1)d
• S{n} = (n/2){ 2a + (n-1)d }

Solution :

Formula of nth term of AP , S{n} = (n/2){ 2a + (n-1)d }

$\sf \implies \: S{n} = (\dfrac{n}{2})(2a + \: (n - 1)d) \\ \\ \sf \implies \: S{n} = (\dfrac{n}{2} )(2a + \: nd - d) \\ \\ \sf \implies \: S{n} = (\dfrac{n}{2} )((2a - d) + \: nd)$

Also we are given that , S{n} = (n/2)(3n + 37)

equating S{n} , from both equation,

$\sf \implies \: (\dfrac{n}{2} )((2a - d) + \: nd) = \dfrac{n}{2} (3n + 37) \\ \\ \sf \dfrac{n}{2} \: get \: cncelled \: from \: both \: side \: we \: get \\ \\ \sf \implies \: ( \: (2a - d) \: + nd \: ) = 37 \: + \: 3n$

Here,

On LHS ,

• "(2a-d)" is constant part
• "nd" is variable part

On RHS

• "37" is constant part
• "3n" is variable part

On comparing variable part on both side together and constant part together,

• nd = 3n
• d = 3n/n
• d = 3

• (2a-d) = 37

putting value of d from above we get,

• 2a - (3) = 37
• 2a = 37+3 = 40
• a = 40/2
• a = 20

Now , we are given that ; a{p} = 56

=> a + (p-1)d = 56

putting value of a & d , we get;

=> 20 + (p-1)3 = 56

=> (p-1)3 = 56-20 = 36

=> p-1 = 36/3

=> p-1 = 12

=> p = 12+1

=> p = 13

ANSWER : Option c) p = 13

Answer 2

Step-by-step explanation:

charu

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read the textbook thoroughly

nd when you cover each chapter along withit do workout previous year question(oswaal q bank is the best)

Nd model papers are avilable on cbse site

This is the best option in scoring good marks in science and maths

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