Which of the following matchings is true for | vedclass

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(A) The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.$1$. For $A.P.$ $1, 3, 5, 7, \ldots$,$a=1, d=2$. The $6^{th}$ term is $a_6 = 1 + (6-

Vedclass · Accepted Answer

$(1-b), (2-c), (3-d), (4-a)$

Vedclass · Answer

(A) The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.$1$. For $A.P.$ $1, 3, 5, 7, \ldots$,$a=1, d=2$. The $6^{th}$ term is $a_6 = 1 + (6-1)2 = 1 + 10 = 11$. So,$(1-b)$.$2$. For $A.P.$ $3, 6, 9, 12, \ldots$,$a=3, d=3$. The $11^{th}$ term is $a_{11} = 3 + (11-1)3 = 3 + 30 = 33$. So,$(2-c)$.$3$. For $A.P.$ $4, 6, 8, 10, \ldots$,$a=4, d=2$. The $16^{th}$ term is $a_{16} = 4 + (16-1)2 = 4 + 30 = 34$. So,$(3-d)$.$4$. For $A.P.$ $5, 10, 15, 20, \ldots$,$a=5, d=5$. The $21^{st}$ term is $a_{21} = 5 + (21-1)5 = 5 + 100 = 105$. So,$(4-a)$.Thus,the correct matching is $(1-b), (2-c), (3-d), (4-a)$.

Part $I$	Part $II$
$1$. Sixth term of the $A.P.$ $1, 3, 5, 7, \ldots$	$a$. $105$
$2$. Eleventh term of the $A.P.$ $3, 6, 9, 12, \ldots$	$b$. $11$
$3$. Sixteenth term of the $A.P.$ $4, 6, 8, 10, \ldots$	$c$. $33$
$4$. Twenty-first term of the $A.P.$ $5, 10, 15, 20, \ldots$	$d$. $34$

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