Question

2. The first and last term of an AP are a andl and sum of the AP is S, then the common
12 - a? Here k is equal to
difference is 7- (1
)
(a) s
(b) 25
(c) 3S
(d) None of these​

Answer 1

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Given : First term of the AP = a

Given : Last term of the AP = l.

Given : Total number of terms = n.

Common difference from the beginning = d.

Common difference from the end = -d.

∴ Last term of the AP = nth term of the AP

⇒ an = l

⇒ a + (n - 1) * d = l

Now,

⇒ nth term of the AP from the beginning an = a + (n - 1) * d

⇒ nth term from the end = l + (n - 1) * (-d)

                                         = l - (n - 1) * d

Sum of nth term from the beginning + nth term from the end is given by:

⇒ a + (n - 1) * d + [l - (n - 1) * d]

⇒ a + (n - 1) * d + l - (n - 1) * d

⇒ a + nd - d + l - nd + d

⇒ a + l.

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Step-by-step explanation:

Given : First term of the AP = a

Given : Last term of the AP = l.

Given : Total number of terms = n.

Common difference from the beginning = d.

Common difference from the end = -d.

∴ Last term of the AP = nth term of the AP

⇒ an = l

⇒ a + (n - 1) * d = l

Now,

⇒ nth term of the AP from the beginning an = a + (n - 1) * d

⇒ nth term from the end = l + (n - 1) * (-d)

= l - (n - 1) * d

Sum of nth term from the beginning + nth term from the end is given by:

⇒ a + (n - 1) * d + [l - (n - 1) * d]

⇒ a + (n - 1) * d + l - (n - 1) * d

⇒ a + nd - d + l - nd + d

⇒ a + l.