Question

Find the mth term of an Arithmetic progression whose 12th

term exceeds the 5th term by 14 and the sum of both terms

is 36​

Answer 1

Answer:

12th term=a+11d

5th term=a+4d

acc to ques.

(a+11d)-(a+4d)=14

7d=14

d=2

also (a+11d)+(a+4d)=36

putting the value of d and solving for a we get

a=3

mth term=a+(m-1)d

=3+(m-1)2

mth term=1+2m

Answer 2

Answer:

QUESTION :

• Find the mth term of an Arithmetic progression whose 12th term exceeds the 5th term by 14 and the sum of both terms is 36

______________

WHAT IS AP ?

• AP or ARITHMETIC PROGRESSION is a systematic concept in maths in which 2 any consecutive terms have same common difference
• The common difference is denoted by d and first term by a

______________

WE KNOW :

Tn in any AP =

$a + (n - 1) \times d$

_____________

SOLUTION:

T12 —T5 =14

• a+11d —a-4d =14
• 7d = 14
• d=2

COMMON DIFFERENCE = 2

T12 +T5 = 36

a+11d+a+4d = 36

but d=2

• 2a +15×2 =36
• 2a = 6
• a=3

$ap \: is \: 3 \\ 5 \\ 7 \\ .......$

_____________

ANSWER :

3,5,7........

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