Step-by-step explanation:
[tex] a_{10} = - 27... (Given) \\\\
\therefore a + 9d = - 27...... (1)\\\\
a_{5} = - 12... (Given) \\\\
\therefore a + 4d = - 12.......(2)\\\\
[/tex]
Subtracting equation (2) from equation (1)
[tex]a + 9d-( a + 4d )= - 27-(-12)\\\\
\therefore a +9d - a-4d = - 27+12\\\\
\therefore 5d = - 15\\\\
\therefore d = \frac {- 15}{5}\\\\
\huge \orange {\boxed {\therefore d = -3}} \\\\
[/tex]
Substituting d = - 3 in equation (1), we find:
[tex]a + 9\times (-3)= - 27\\\\
\therefore a - 27 = - 27\\\\
\therefore a = 27- 27\\\\
\huge \red {\boxed {\therefore a = 0}}\\\\[/tex]
Hence, first term is zero.
Information provided with us :
What we have to calculate :
Performing Calculations :
Now, as we clearly know that nth or general term of an A.P. (Arithmetic progression) s calculated by the formula :
Here in this formula,
★ For 10th term of the A.P. :
=> a + (10 - 1) d = -27
=> a + (9) d = -27
=> a + 9 × d = -27
=> a + 9d = -27
=> a = -27 - 9d
Here we got a temporary value of first term.
★ For 5th term of the A.P. :
=> a + (5 - 1) d = -12
=> a + 4d = -12
Substituting the value of a which we got above,
=> -27 - 9d + 4d = -12
=> -27 - 5d = -12
=> -5d = -12 + 27
=> -5d = 15
=> -d = 15 / 5
=> -d = 3
=> d = -3
Therefore, common difference (d) is -3.
★ Finding out first term of A.P. :
=> a = -27 - 9d
=> a = -27 - 9 (-3)
=> a = -27 - 9 × -3
=> a = -27 - 27
=> a = 0
★ Henceforth,
