Question

the 2 term and the 21st term of an a. p. are 16 and 29 respectively find the 41st term of this a. p.

Answer 1

Answer:


t­11 = 16   and t21 =  29    [Given]


We know that,


tn =  a + (n – 1 )d

∴ t11 = a + (11 – 1 )d

∴ 16 = a + 10d    ------ eq. no. 1


∴ t21 = a + (21 – 1) d

∴ 29 = a + 20 d ---------eq. no. 2


Subtracting eq. 1 from eq. 2


29 = a + 20 d

16 = a + 10d   

-           -    -  

13 = 0 + 10d


∴ 13 = 10d

∴ 13/10 = d

∴ 1.3 = d


Substituting d = 1.3 in equation 1


∴ 16 = a + 10d

∴ 16 = a + 10(1.3)

∴ 16 = a + 13

∴ 16 – 13 = a

∴ 3 = a


To find the 34th Term


t34 = a + (34 – 1 )d

t34 = 3 + 33 (1.3)

t34 = 3  + 42.9

t34 = 45.9


To find .   ‘n’ such that tn = 55


tn = a + (n – 1) d

∴ 55 = 3 + (n – 1) 1.3

∴ 55 – 3 = 1.3 (n – 1)

∴ 52 = 1.3 (n – 1)

∴ 52/1.3 = (n – 1)

∴ 40 = n – 1

∴ 40 + 1 = n


∴ n = 41