Question

find the first term of A.P.whose second term is 14 and common difference is 14.​

Answer 1

To Find :

• we need to find the 1st term of AP.

SOLUTION :

• Second term of AP = 14
• Common difference = 14

we know that,

$\bf \: a_n = a +( n- 1) d$

where ,

• an = nth term
• n = number of terms
• a = first term
• d = common difference

↱ 2nd term = a + d

⟶ a2 = a + (2 - 1)d

⟶ a2 = a + d

⟶ a + d = 14

⟶ a + 14 = 14

⟶ a = 14 - 14

⟶ a = 0

Verification : -

⟹ a + d = 14

⟹ 0 + 14 = 14

⟹ 14 = 14

LHS = RHS

hence Verified

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Answer 2

$\sf \huge \green {question \: : }$

Find the first term of A.P.whose second term is 14 and common difference is 14.

$\sf \huge \green{answer \: : }$

a = 0

$\bf \huge\red{given}$

$\bf \huge \star \: second \: term \: = 14$

$\bf \huge \star \: common \: difference \: = 14$

$\bf \purple{ \: formula \: used \: \: }$

$\bf \huge \star \fbox \pink{ \: \: \: a_n = a + (n - 1)d \: \: }$

$\sf \huge \: solution :$

$\bf \huge \implies \: a_n \: = a + (n - 1)d$

$\bf \huge \implies \: a_2 \: =a + (2 - 1)14$

$\bf \huge \implies \: 14 \: =a + 14$

$\bf \huge \implies \: a + 14 \: =14$

$\bf \huge \implies \: a\: =14 - 14$

$\bf \huge \implies \: a \: =0$

$\bf \red{ \huge \star } \huge\fbox \green{ \: \: a = 0 \: \: }$