Question

In an A.P. if S11=77 , find T6.

Answer 1

Answer:

t6 = 45

but, tn = a+(n-1)d

t6.= a +5d

45 = a+5d

a= 45-5d

t11 = a+10d

Sn = n/2 {2a+(n-1)d}

S11 = 11/2 {2×(45-5d) + 10d

S11 = 11/2 {90-10d +10d}

S11 = 11/2 × 90

S11 = 11×45

S11 = 495

Answer 2

Answer:

The 6th term of Arithmetic Progression is, T₆ = 7

Step-by-step explanation:

Arithmetic Progression:

An arithmetic sequence or progression is defined as a sequence of numbers in which the difference between two consecutive terms is always a fixed constant. This fixed constant that is added to any term to get the next term in the sequence is known as Common Difference (d). The first term in the sequence is denoted by 'a' and the number of terms is denoted by 'n'. Then, the arithmetic sequence can be written as

a, a+d, a+2d, a+3d, ......., a+(n-1)d

General form or nth term of an Arithmetic sequence:

Tₙ = a + (n - 1) d

Sum of n terms of an Arithmetic Progression

The sum of the first n terms of an arithmetic progression can be found using the formula,

Sₙ = n/2 [ 2a + (n - 1) d ]

where n ⇒ number of terms

           a ⇒ first term

           d ⇒ common difference

Given:

Sum of first 11 terms, S₁₁ = 77

Find:

The $6 \sp t\sp h$ term, T₆ = ?

Solution:

Given that, S₁₁ = 77

⇒ $\frac{11}{2} [ \ 2a + (11-1)d \ ] = 77$            (∵ Sₙ = $\frac{n}{2} [ \ 2a+ ( n-1)d \ ]$ )

⇒ $\frac{11}{2} [ \ 2a + 10d \ ] = 77$

⇒ $11 [ \ 2a+10d \ ] = 77 \times 2$

⇒ $11 [ \ 2a+10d \ ] = 154$

⇒ $2a+10d = \frac{154}{11}$

⇒ $2a+10d = 14$

⇒ $2 \ (a+5d) = 2 \times 7$                     ( Taking 2 in common )

⇒ $a+5d = 7$

We know that, $a+5d$ is in the form of $a+(n-1)d$

Here, $n-1 =5 \implies n=6$

i.e., 6th term, T₆ = 7

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