Question

in an a.p an+5=35 and an+1=23 then common difference is?​

Answer 1

→ $\mathsf{a_n}$ + 5 = 35

→ a + (n + 5 - 1)d = 35

→ a + (n + 4)d = 35

→ a + nd + 4d = 35 __(i)

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→ $\mathsf{a_n}$ + 1 = 23

→ a + (n + 1 - 1)d = 23

→ a + nd = 23 __(ii)

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By substituting (ii) in (i),

→ 23 + 4d = 35

→ 4d = 35 - 23

→ d = 3

$\huge{\boxed{\mathsf{Comman\:d : 3}}}$

Answer 2

Answer:-

Equation 1

$an + 5 = 35 \\ </p><p>a + (n + 5 - 1)d = 35 \\ </p><p>a + (n + 4)d = 35 \\ </p><p>a + nd + 4d = 35 \\ </p><p>$

Equation 2

$</p><p> an + 1 = 23 \\ </p><p>a + (n + 1 - 1)d = 23 \\ </p><p> a + nd = 23$

Now next we should substitute both equation a and b

$23 + 4d = 35 \\ </p><p> 4d = 35 - 23 \\ </p><p> difference \: = 3</p><p>$

Answer=3