Question

3rd term of AP is 18
and
8th termi's 48 3
find the first s terms of A
& ind
12th term
find algebraic expresion
for given AP​

Answer 1

Let the first term of an AP be a and common difference be d

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12Therefore sum of first 17 terms in AP is

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12Therefore sum of first 17 terms in AP is 2

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12Therefore sum of first 17 terms in AP is 217

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12Therefore sum of first 17 terms in AP is 217 (2×12+(17−1)×3)=

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12Therefore sum of first 17 terms in AP is 217 (2×12+(17−1)×3)= 2

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12Therefore sum of first 17 terms in AP is 217 (2×12+(17−1)×3)= 217

Let the first term of an AP be a and common difference be dThe third term of an AP is a+2d, given that it is equal to 18⇒a+2d=18The seventh term of an AP is a+6d, given that it is equal to 30⇒a+6d=30By subtracting first equation from second equation , we get 4d=12⇒d=3By substituting the value of d it in first equation, we get a=12Therefore sum of first 17 terms in AP is 217 (2×12+(17−1)×3)= 217 ×72=612